Final answer:
The force exerted by rope R1 on block 1 is equal to the tension in that rope, and if both blocks have the same acceleration and there are no other external forces, this tension is also equal to the force R1 exerts on block 2. However, the force applied through rope R2 on block 2 may be larger because it must account for the total mass and acceleration of both blocks.
Explanation:
When two blocks are connected by a rope and a force is applied to one of the blocks (via a second rope), the tensions in the ropes are subjects of interest in Physics, specifically classical mechanics. In this scenario, the force applied to rope R2 is used to accelerate both blocks 1 and 2. The tension in rope R1, which connects blocks 1 and 2, should be analyzed using a free-body diagram for each block.
Rope R1 exerts a force on block 1, and rope R2 exerts a force on block 2. The force exerted by rope R2 must be sufficient not only to accelerate block 2 but also to account for the force needed to accelerate block 1. Therefore, the force exerted by R2 on block 2 is equal to the sum of the forces necessary to accelerate both blocks. This total force must be greater than or equal to the force that R1 exerts on block 1, as rope R1 only has to provide the force needed to accelerate block 1. However, if we assume the acceleration of the two blocks is the same and the system is not experiencing any external forces other than the applied force, the tension in rope R1 will be equal to the force R1 exerts on block 1 because the massless ropes can only transmit the force equally along their lengths.
Therefore, it can be concluded that the magnitude of the force exerted by rope R1 on block 1 is equal to the magnitude of the tension on block 2 by rope R1, but the magnitude of the applied force through R2 on block 2 could be larger as it has to account for the combined mass of block 1 and block 2.
Erica (39kg \;kg ) and Danny (45kg \;kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.8m/s \; m/s . At that instant he grabs hold of her.
What is their speed just after he grabs her?
help please
can you show formula too
Answer:
2.57 m/s
Explanation:
mass of Erica (ME) = 39 kg
velocity of Erica (VE) = 0 m/s (since she is at her high point of her jump, her velocity will be 0)
mass of Danny (MD) = 45 kg
velocity of Danny (VD) = 4.8 m/s
from the conservation of momentum total initial momentum is equal to the total final momentum
total initial momentum = final initial momentum
(ME)(VE) + (MD)(VD) = (ME + MD) V
where "V" is the velocity of Erica and Danny after Danny grabs her.
(39 x 0) + (45 x 4.8) = (39 + 45) x V
0 + 216 = 84 V
V = 216 / 84 = 2.57 m/s
The speed just after Danny grabs Erica is equal to 2.57 m/s.
Given the following data:
Mass of Erica = 39 kgMass of Danny = 45 kgVelocity of Erica = 0 m/s (since she is at the highest point)Velocity of Danny = 4.8 m/sTo determine their speed just after Danny grabs Erica, we would apply the law of conservation of momentum:
[tex]M_EV_E + M_DV_D = (M_E + M_D)V_{f}[/tex]
Where:
[tex]M_E[/tex] is the mass of Erica.[tex]M_D[/tex] is the mass of Danny.[tex]V_f[/tex] is the final speed.[tex]V_E[/tex] is the speed of the Erica.[tex]V_D[/tex] is the speed of the Danny.Substituting the given parameters into the formula, we have;
[tex]39 \times 0 + 45 \times 4.8 = (39 +45)V_f\\\\0+216=84V_f\\\\V_f = \frac{216}{84} \\\\V_f = 2.57 \;m/s[/tex]
Final speed = 2.57 m/s
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a beam containing a mixture of helium isotopes enters a mass spectrometer. The beam then divides into two beams that each travel to a different point on the detector. Which property do the particles in the two beams have in common?
A. each particle has the same mass
B. Each particle is a helium atom, not an ion
C. Each particle contains the same number of neutrons
D. Each particle has the same initial velocity
Answer: did you get the answer(s) by chance?
Explanation:
Answer:
The last one
Explanation:
The decibel level of an orchestra is 90 db, and the violin section achieves a level of 80 dB. How does the sound intensity from the full orchestra compare to that from the violin section alone?
Final answer:
The sound intensity of the full orchestra is ten times greater than that of the violin section alone because a 10 dB increase corresponds to a tenfold increase in intensity.
Explanation:
The decibel level (dB) measures sound intensity on a logarithmic scale, where each 10 dB increase represents a tenfold increase in intensity. To compare the sound intensity of the full orchestra at 90 dB to the violin section at 80 dB, we can use the fact that an increase of 10 dB corresponds to a tenfold increase in intensity. Therefore, the full orchestra's sound intensity is ten times greater than that of the violin section alone.
This simplifies to 10, meaning that the sound intensity from the full orchestra is 10 times greater than that from the violin section alone.
In simpler terms, when comparing sound intensities, every 10 dB increase corresponds to a tenfold increase in intensity. Therefore, the difference of 10 dB between the full orchestra and the violin section results in the orchestra being 10 times louder in terms of sound intensity.
The sound intensity from the full orchestra is [tex]\( {10} \)[/tex] times greater than the sound intensity from the violin section alone.
To compare the sound intensities of the full orchestra and the violin section, we need to understand the relationship between decibels (dB) and sound intensity [tex]\( I \)[/tex].
The decibel scale is logarithmic and relates to sound intensity [tex]\( I \)[/tex] in the following way:
[tex]\[ \text{dB} = 10 \log_{10} \left( \frac{I}{I_0} \right) \][/tex]
where [tex]\( I \)[/tex] is the sound intensity of interest and [tex]\( I_0 \)[/tex] is the reference intensity (typically [tex]\( I_0 = 10^{-12} \)[/tex] W/m[tex]\(^2\))[/tex].
Step 1: Convert dB to intensity ratio
Given:
- Orchestra sound level [tex]\( L_{\text{orchestra}} = 90 \)[/tex] dB
- Violin section sound level [tex]\( L_{\text{violin}} = 80 \)[/tex] dB
The difference in decibel levels between the orchestra and the violin section gives us the intensity ratio:
[tex]\[ L_{\text{orchestra}} - L_{\text{violin}} = 90 \, \text{dB} - 80 \, \text{dB} = 10 \, \text{dB} \][/tex]
The intensity ratio in terms of decibels is related by:
[tex]\[ 10 \log_{10} \left( \frac{I_{\text{orchestra}}}{I_0} \right) - 10 \log_{10} \left( \frac{I_{\text{violin}}}{I_0} \right) = 10 \][/tex]
Step 2: Calculate the ratio of sound intensities
To find [tex]\( \frac{I_{\text{orchestra}}}{I_{\text{violin}}} \)[/tex]:
[tex]\[ \frac{I_{\text{orchestra}}}{I_{\text{violin}}} = 10^{10 / 10} \][/tex]
[tex]\[ \frac{I_{\text{orchestra}}}{I_{\text{violin}}} = 10^{1} \][/tex]
[tex]\[ \frac{I_{\text{orchestra}}}{I_{\text{violin}}} = 10 \][/tex]
Sound is more effectively transmitted into a stethoscope by direct contact than through the air, and it is further intensified by being concentrated on the smaller area of the eardrum. It is reasonable to assume that sound is transmitted into a stethoscope 100 times as effectively compared with transmission though the air. What, then, is the gain in decibels produced by a stethoscope that has a sound gathering area of 15.0 cm2, and concentrates the sound onto two eardrums with a total area of 1.00 cm2 with an efficiency of 37.0%?
Answer:
Δβ = 28.2 dB
Explanation:
Attached is the explanation
The height of a typical playground slide is about 6 ft and it rises at an angle of 30 ∘ above the horizontal.
a.)Some children like to slide down while sitting on a sheet of wax paper. This makes the friction force exerted by the slide very small. If a child starts from rest and we take the friction force to be zero, what is the speed of the child when he reaches the bottom of the slide?
b.)If the child doesn't use the wax paper, his speed at the bottom is half the value calculated in part A. What is the coefficient of kinetic friction between the child and the slide when wax paper isn't used? μk = ?
I found the answer to part a, it is 3m/s, but I need part b. Please show all steps, thank you!
Answer:
What is the coefficient of kinetic friction = 0.432
Explanation:
The detailed steps and derivation with appropriate substitution is as shown in the attached file.
This question involves the concepts of the law of conservation of energy and frictional energy.
a) The speed of the child when he reaches the bottom of the slide is "6 m/s".
b) The coefficient of kinetic friction between the child and slide when the wax paper isn't used is "0.432".
a)
According to the law of conservation of energy in this situation:
Loss in Potential Energy = Gain in Kinetic Energy
[tex]mgh = \frac{1}{2}mv^2\\\\2gh=v^2\\v=\sqrt{2gh}\\\\[/tex]
where,
v = velocity = ?
g = acceleration due to gravity = 9.81 m/s²
h = height lost = 6 ft = 1.83 m
Therefore,
[tex]v=\sqrt{2(9.81\ m/s^2)(1.83\ m)}[/tex]
v = 6 m/s
b)
Now, the velocity becomes half and the friction comes into action. So in this case the law of conservation of energy will be written as:
Loss of Potential Energy = Gain of Kinetic Energy + Frictional Energy
[tex]mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kRl\\\\mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kmgCos\theta l\\\\gh-\frac{1}{2}(\frac{v}{2})^2+\mu_kgCos\theta l[/tex]
where,
l = length of slide = [tex]\frac{h}{sin\theta}=\frac{1.83\ m}{sin30^o}=3.66\ m[/tex]
[tex]\mu_k[/tex] = coefficient of kinetic friction = ?
Therefore,
[tex](9.81\ m/s^2)(1.83\ m)-\frac{1}{2}(\frac{6\ m/s}{2})^2=\mu_k(9.81\ m/s^2)(Cos30^o)(3.66\ m)\\\\\mu_k=\frac{13.45\ m^2/s^2}{31.09\ m^2/s^2}\\\\\mu_k=0.432[/tex]
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The attached picture explains the law of conservation of energy.
Railroad cars are loosely coupled so that there is a noticeable time delay from the time the first and last car is moved from rest by the locomotive. Discuss the advisability of this loose coupling and slack between cars from the point of view of impulse and momentum.
Answer:
Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well).
Explanation:
After takeoff, an airplane climbs at an angle of 30° at a speed of 215 ft/sec. How long does it take for the airplane to reach an altitude of 13,000 ft? (Round your answer to one decimal place.)
Answer:
120.9 seconds
Explanation:
*Attached below is a rough sketch of the problem. Point A represents the point where the airplane climbs 30°, BC represents the altitude of the airplane, AC represents the angular displacement of the airplane.
Parameters given:
Angle of elevation = 30°
Speed (angular) of the airplane = 215 ft/sec
Altitude = 13000 ft
We need to find the angular displacement of the airplane to find the time it takes to get to an altitude of 13000 ft. The angular displacement is represented by the hypotenuse of the triangle. Hence, using SOHCAHTOA,
[tex]sin30^{o} = \frac{13000}{hyp} \\\\hyp = \frac{13000}{sin30^{o}}\\ \\hyp = 26000 ft[/tex]
∴ time taken = [tex]\frac{angular displacement}{angular speed}\\ \\[/tex]
[tex]time taken = \frac{26000}{215} \\\\time taken = 120.9 seconds[/tex]
Therefore, it will take the airplane 120.9 seconds to get to an altitude of 13000 ft.
According to the cartoon video on physics, electrons sent through 2 slits at once without measuring what slit is goes through, will produce what?
A) An interference pattern
B) A single line
C) 4 lines
D) A blank screen
Answer:
A) An interference pattern
Explanation:
the two slit experiment is key to understand the microscopic world. The wave-like properties of light were demonstrated by the famous experiment first performed by Thomas Young in the early nineteenth century. In original experiment, a point source of light illuminates two narrow adjacent slits in a screen, and the image of the light that passes through the slits is observed on a second screen.
Key Points
waves can interfere, for light this will make a series of light and dark bands matter particles, such as electrons, also produce interference patterns due to their wave-like nature so with a high flux of either photons or electrons, the characteristic interference pattern is visibleA localized electron has been polarized so that its spin is oriented in the positive z- direction. It is now subject to the application of a constant uniform magnetic field B = Bî along x over a period of time of duration T. After that, it is subject to the application of another magnetic field of the same magnitude B but along y: B2 = Bý, also with duration T. (a) What is the probability P that the spin-flip would occur as a result? That is, what is the probability that the spin of the electron would be found oriented in the negative z-direction after the application of the magnetic fields is over? (b) Is it possible to find such duration that the spin-flip would occur with certainty: P = 1? If yes, what would be that time r?
Answer: The answer is attached
Explanation:
Final answer:
In quantum mechanics, the probability of a spin-flip and the potential for achieving a spin flip with certainty are explored through the application of magnetic fields in different directions.
Explanation:
P(a): The probability that the spin-flip would occur can be calculated using quantum mechanics. When applying magnetic fields B along x and y, the probability of finding the spin of the electron oriented in the negative z-direction after is over is determined by the transition probabilities.
P(b): The time to ensure a spin flip with certainty (P = 1) can be calculated by adjusting the duration of the magnetic field application. By manipulating the duration, the spin can be controlled to achieve a certain outcome at a specific time.
A flying dragon is rising vertically at a constant speed of 6.0m/s. When the dragon is 30.0m above the ground, the rider on its back drops a small golden egg which, subsequently, is in free fall.
a) What is the maximum height above the ground reached by the egg?
b) How long after its release does the egg hit the ground?
c) What is the egg’s velocity immediately before it hits the ground?
d) Sketch, qualitatively, position, velocity, and acceleration of the egg as functions of time.
Final answer:
The detailed response covers the maximum height reached by the egg, the time it takes to hit the ground, and its velocity just before landing.a)60.0m,b)2.45 seconds,c)-6.0m/s.
Explanation:
a) Maximum Height: The maximum height above the ground reached by the egg can be calculated using the kinematic equation. It will be twice the initial height. In this case, it would be 60.0m.
b) Time to Hit Ground: You can determine the time it takes for the egg to hit the ground by using the kinematic equation for vertical motion. The time would be approximately 2.45 seconds after its release.
c) Velocity Before Landing: The velocity of the egg immediately before hitting the ground would be the same as the initial velocity when it was thrown, but in the opposite direction, which is -6.0m/s.
If an astronaut goes on a space walk outside the Space Station, she will quickly float away from the station unless she has a tether holding her to the station. Part A Choose the correct explanation why does the statement make sense (or is clearly true) or does not make sense (or is clearly false). Choose the correct explanation why does the statement make sense (or is clearly true) or does not make sense (or is clearly false).
a. This statement is true. She and the Space Station have different orbits at the beginning and will move apart.
b. This statement is true. She and the Space Station cannot share the same orbit and will move apart quickly.
c. This statement is false. She and the Space Station have different orbits at the beginning but will stay together due to mutual gravity.
d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.
Answer:
d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.
Explanation:
In astronomy, orbit is simply a path of an object around another object in a space. That is, orbit is a path of a body that revolves around a gravitating center of mass. Examples of an orbit is are satellite around a planet, orbit around a center of galaxy, planet around the sun, and among others.
On the other hand, space station refers to a spacecraft that can support a group of human for long time in the orbit. Another names for space stations are orbital space station and orbital station.
Therefore, an astronaut goes on a space walk outside the Space Station shares the same orbit with the space station and they will stay together unless they are pushed apart.
The statement is true that an astronaut would float away during a spacewalk if not tethered. The astronaut and ISS, although influencing each other gravitationally to a small extent, they would continue along their separate paths under the influence of Earth's gravity and microgravity.
Explanation:
This statement is true. When an astronaut steps out for a spacewalk, she and the International Space Station (ISS) are both in orbit around Earth but the astronaut will not stay in place relative to the ISS without a tether holding her to the station.
The reason for this is due to microgravity and the properties of motion in space. The astronaut and the ISS are both in free-fall around the Earth, so in the absence of other forces, they would continue along their separate paths. Without a tether, even a small force (like the push off the astronaut does to get away from the airlock) can cause her to drift away from the station.
While the ISS and the astronaut do influence each other gravitationally, this effect is extremely small compared to the force of Earth's gravity. So, without a tether, the astronaut can float away from the ISS.
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An elevator packed with passengers has a mass of 1950 kg.
(a) The elevator accelerates upward (in the positive direction) from rest at a rate of 2 m/s2 for 2.05 s. Calculate the tension in the cable supporting the elevator in newtons.
(b) The elevator continues upward at constant velocity for 8 s. What is the tension in the cable, in Newtons, during this time?
(c) The elevator experiences a negative acceleration at a rate of 0.45 m/s2 for 2.6 s. What is the tension in the cable, in Newtons, during this period of negative accleration?
(d) How far, in meters, has the elevator moved above its original starting point?
Answer:
a) Fa= Tension in the cable= 23010N
b) 19110N
c) 18232.5N
d=41.28m ,final velocity=0
Explanation:
a) Newtons 2nd law is given by Fnet=EF=ma
Fa= m(a + g)
Fa= 1950 (2+9.8)
Fa= 1950×11.8= 23010N
b) Fnet=0
Therefore Fb= W= 1950×9.8= 19110N
c) Fc= m(g - a)
Fc= 1950(9.8 - 0.45)
Fc= 18232.5N
d) First distance
ya= vt + 0.5at^2 = 0 + (0.5)(2)(2.05)^2
ya= 4.203m
yb= vt= 4.203×8=33.62m
yc = vt - (0.5at^2)
yc= 4.203×2.6) - (0.5×0.45×8^2)
yc = 10.93-14.4
yc =-3.46m
Dtotal = -3.46+33.62+4.203
Dtotal=41.28m
Explanation:
Below is an attachment of the solution.
A light ray passes from air through a glass plate with refractive index 1.60 into water. The angle of the refracted ray in the water is 42.0°. Determine the angle of the incident ray at the air-glass interface?
Answer:
The angle of the incident ray at the air-glass interface is 62.86°
Explanation:
Given that,
Refractive index of glass =1.60
Angle = 42°
We need to calculate the angle of the incident ray at glass-water interface
Using Snell's law
[tex]n_{2}\sin\theta_{2}=n_{3}\sin\theta_{3}[/tex]
Where, [tex]n_{2}[/tex] = refractive index of glass
[tex]n_{3}[/tex] = refractive index of water
[tex]\theta_{3}[/tex] = angle of refraction
Put the value into the formula
[tex]1.6\sin\theta_{2}=1.33\sin42[/tex]
[tex]\sin\theta_{2}=\dfrac{1.33\sin42}{1.6}[/tex]
[tex]\theta_{2}=\sin^{-1}(\dfrac{1.33\sin42}{1.6})[/tex]
[tex]\theta_{2}=33.8^{\circ}[/tex]
We need to calculate the angle of the incident ray at the air-glass interface
Using Snell's law
[tex]n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}[/tex]
Where, [tex]n_{1}[/tex] = refractive index of air
[tex]n_{2}[/tex] = refractive index of glass
[tex]\theta_{1}[/tex] = angle of incident
Put the value into the formula
[tex]1\sin\theta_{1}=1.6\sin33.8[/tex]
[tex]\theta=\sin^{-1}(1.6\sin33.8)[/tex]
[tex]\theta=62.86^{\circ}[/tex]
Hence, The angle of the incident ray at the air-glass interface is 62.86°
Answer:
62.8 degree
Explanation:
Let the incident ray incident at an angle [tex]\theta_1[/tex] at air glass surface.
[tex]\theta_3=42^{\circ}[/tex]=Angle of refraction when ray travel from glass to water
[tex]\theta_2=[/tex]Angle of refraction when the ray travel from air to glass
Refractive index of glass,[tex]n_2=1.6[/tex]
We know that
Refractive index of water=[tex]n_3=1.33[/tex]
Snell's law
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
Where [tex]\theta_1[/tex]=Angle of incidence
[tex]\theta_2=[/tex]Angle of refraction
[tex]n_1=[/tex]Refractive index of medium 1
[tex]n_2=[/tex]Refractive index of medium 2
When the ray travel from glass to water
[tex]n_2sin\theta_2=n_3sin\theta_3[/tex]
Where [tex]n_2=[/tex]Refractive index of medium 1(Glass)
[tex]n_3[/tex]=Refractive index of medium 2 (Water)
[tex]\theta_2=[/tex]Angle of incidence
[tex]\theta_3[/tex]=Angle of refraction
Substitute the values
[tex]1.6sin\theta_2=1.33sin42[/tex]
[tex]sin\theta_2=\frac{1.33sin42}{1.6}[/tex]
[tex]sin\theta_2=0.556[/tex]
[tex]\theta_2=sin^{-1}(0.556)=33.8^{\circ}[/tex]
Angle of refraction when ray travel from air to glass= Angle of incidence of when ray travel from glass to water
Angle of refraction when the ray travel from air to glass=33.8 degree
Refractive index of air=[tex]n_1=1[/tex]
Again apply Snell's law
[tex]n_1sin\thet_1=n_2sin\theta_2[/tex]
[tex]1\times sin\theta_1=1.6sin(33.8)[/tex]
[tex]sin\theta_1=1.6\times 0.556=0.8896[/tex]
[tex]\theta_1=sin^{-1}(0.8896)=62.8^{\circ}[/tex]
Hence, the angle of the incident ray at the air-glass interface=62.8 degree
A 100-watt electric incandescent light bulb consumes __________ J of energy in 24 hours. [1 Watt (W)
Answer : The energy consumed by bulb in 24 hours is, 8.64 × 10⁶ J
Explanation :
As we are given that:
1 watt = 1 J/s
So,
100 watt = 100 J/s
Now we have to calculate the energy consumed by bulb in 24 hours.
As we know that:
1 hr = 60 min
1 min= 60 sec
So,
24 hr = 24 × 60 × 60 sec = 86400 sec
As, the energy consumed by bulb in 1 second = 100 J
So, the energy consumed by bulb in 86400 second = 86400 × 100 J
= 8.64 × 10⁶ J
Thus, the energy consumed by bulb in 24 hours is, 8.64 × 10⁶ J
To calculate the energy consumed by a 100-watt electric incandescent light bulb in 24 hours, we use the formula E = Pt, where E is the energy in joules (J), P is the power in watts (W), and t is the time in seconds (s). In this case, the energy consumed is 8640000 J.
Explanation:To calculate the electrical energy used by a 100-watt electric incandescent light bulb in 24 hours, we can use the formula E = Pt, where E is the energy in joules (J), P is the power in watts (W), and t is the time in seconds (s).
First, we need to convert the time from hours to seconds. There are 3600 seconds in an hour. So, 24 hours is equal to 24 x 3600 = 86400 seconds.
Now, we can substitute the values into the formula. E = 100W x 86400s = 8640000 J.
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The liquid in one container drops 100 F, while the same liquid in a different container drops 100C. How does the change in thermal energy in the two compare? Explain your answer.
Answer:
Explanation:
The thermal energy is given as
E=mc(T2-T1)
For the temperature change.
ΔT can be expressed in units of Kelvin or degrees Celsius. The ΔT of Kelvin is equal to that of ΔT of Celsius but not equal to the ΔT of Fahrenheit.
Therefore change in temperature of 100F is not equal to change in temperature of 100C
°F= 9/5 °C +32
So let assume 20°C, so the increase of 100°C will give 120°C.
Then °F = 68°F
Now the equivalent of 20°C is 68°F.
So let see the value of 120°C, which is the increase given.
°F= 9/5 ×120+32
°F= 248°F
The change in temp in Fahrenheit is 248-68=180°F
1°F change is = to 0.56°C change
Therefore, 100°F change in temperature in Fahrenheit is equal to 56°C in Celsius
Therefore the thermal energy is not the same, since they have are the same mass and specific heat but different changer in temperature.
So using the formulae given above, the thermal energy of the 100F change is greater than that of the 100C change..
A temperature drop of 100°C corresponds to a larger change in thermal energy than a drop of 100°F because 100°C equates to a more substantial temperature change (around 180°F). The container with the 100°C temperature drop, therefore, experiences a greater thermal energy change.
Explanation:In comparing the change in thermal energy of liquids with a temperature drop of 100°F and 100°C, understanding that 1°C is equal to 1.8°F is crucial. To compare the energy change between the two, we must first recognize that the change of 100°F is approximately equivalent to a change of 55.5°C (100°F / 1.8 = 55.5°C).
Since the specific heat capacity is a property of a material that indicates how much heat energy it requires to change its temperature by a single degree, and this value is typically given in J/g°C, it is clear that a larger temperature change in Celsius would require more energy. Therefore, the container experiencing a drop of 100°C will require the removal of more thermal energy compared to the container with a drop of 100°F.
In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to push grocery carts up the ramps and it is obviously desirable that this not be too difficult. The engineer has done a survey and found that almost no one complains if the force directed up the ramp is no more than 20N .
Ignoring friction, at what maximum angle θ should the ramps be built, assuming a full 30kg grocery cart?
Express your answer using two significant figures.
Answer:
3.90 degrees
Explanation:
Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is
W = mg = 30*9.81 = 294.3 N
This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.
We can calculate the parallel one since it's the one that affects the force required to push up
F = WsinΘ
Since customer would not complain if the force is no more than 20N
F = 20
[tex]294.3sin\theta = 20[/tex]
[tex]sin\theta = 20/294.3 = 0.068[/tex]
[tex]\theta = sin^{-1}0.068 = 0.068 rad = 0.068*180/\pi \approx 3.90^0[/tex]
So the ramp cannot be larger than 3.9 degrees
A bullet is fired through a wooden board with a thickness of 9.0 cm. The bullet hits the board perpendicular to it, and with a speed of +370 m/s. The bullet then emerges on the other side of the board with a speed of +259 m/s. Assuming constant acceleration (rather, deceleration!) of the bullet while inside the wooden board, calculate the acceleration.
Remember that since the bullet is traveling in the positive direction and it is slowing down, the acceleration is in the opposite, or negative, direction. (In this case the acceleration is negative.)
Calculate also the total time the bullet is in contact with the board (in sec).
Answer:
-387883.3 m/s²
0.000286168546055 seconds
Explanation:
t = Time taken
u = Initial velocity = 370 m/s
v = Final velocity = 259 m/s
s = Displacement = 9 cm
a = Acceleration
[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{259^2-370^2}{2\times 9\times 10^{-2}}\\\Rightarrow a=-387883.3\ m/s^2[/tex]
The acceleration of the bullet is -387883.3 m/s²
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{259-370}{-387883.3}\\\Rightarrow t=0.000286168546055\ s[/tex]
The time taken is 0.000286168546055 seconds
Final answer:
The acceleration of the bullet is approximately -111 m/s^2 and the total time the bullet is in contact with the board is approximately 0.00081 seconds.
Explanation:
The acceleration of the bullet can be calculated using the equation of motion:
a = (v - u) / t
where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time. In this case, the initial velocity of the bullet is 370 m/s (positive because it is in the direction of motion) and the final velocity is 259 m/s (positive because it is also in the direction of motion). The time it takes for the bullet to decelerate can be calculated using the equation:
x = ut + (1/2)at^2
where x is the displacement, u is the initial velocity, t is the time, and a is the acceleration. In this case, the displacement is the thickness of the wooden board, which is 9.0 cm (or 0.09 m). Rearranging the equation, we get:
t = (v - u) / a
Substituting the values, we have:
t = (259 - 370) / a
Combining the equations for time and acceleration, we get:
t = (0.09) / (370 - 259) = (0.09) / (111) = 0.00081 seconds.
So, the acceleration of the bullet is approximately -111 m/s^2 (negative because it is in the opposite direction of the bullet's motion) and the total time the bullet is in contact with the board is approximately 0.00081 seconds.
Two sinusoidal waves are moving through a medium in the positive x-direction, both having amplitudes of 7.00 cm, a wave number of k=3.00m−1, an angular frequency of ω=2.50s−1, and a period of 6.00 s, but one has a phase shift of an angle ϕ=π12rad. What is the height of the resultant wave at a time t=2.00s and a position x=0.53m?
The question combines the principles of wave mechanics, specifically sinusoidal waves, to calculate the height of a resultant wave at a certain time and position. This is done by applying values to the formula y (x, t) = A sin (kx - wt +φ), where A, k, w and φ stand for amplitude, wave number, angular frequency and phase shift respectively. Applying the values of the moving sinusoidal waves, you can solve for the height of the resultant wave.
Explanation:A sinusoidal wave is represented by the equation y (x, t) = A sin (kx — wt + φ). Here A is the amplitude, k is wave number, w is the angular frequency and φ is the phase shift. Given two sinusoidal waves, moving through a medium, both having the equal amplitude (7.00 cm), wave number (k=3.00m−1) and angular frequency (ω=2.50s−1), but with one having a phase shift of π/12 radians, the resultant wave has an amplitude of AR = [2A cos(φ/2)] and a phase shift equal to half the original phase shift.
Applying these values to the equation y (x, t) = AR sin (kx - wt + φ/2), we have y = 2*7* cos(π/12/2) cm * sin(3.00m−1 *0.53m — 2.50s−1* 2.00s + π/12 /2), thus we can calculate the height of the resultant wave at a particular position and time.
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The height of the resultant wave at [tex]\( t = 2.00 \, \text{s} \) and \( x = 0.53 \, \text{m} \)[/tex] is:
[tex]\[ \boxed{-13.965 \, \text{cm}} \][/tex].
The height of the resultant wave at time [tex]\( t = 2.00 \, \text{s} \)[/tex] and position [tex]\( x = 0.53 \, \text{m} \)[/tex] is given by the sum of the two individual waves, taking into account their respective phase shifts.
The general equation for a sinusoidal wave moving in the positive x-direction is:
[tex]\[ y(x,t) = A \cos(kx - \omega t + \phi) \][/tex]
[tex]where \( A \) is the amplitude, \( k \) is the wave number, \( \omega \) is the angular frequency, \( \phi \) is the phase shift, \( x \) is the position, and \( t \) is the time.[/tex]
For the first wave with no phase shift, the equation is:
[tex]\[ y_1(x,t) = 7.00 \cos(3.00x - 2.50t) \][/tex]
For the second wave with a phase shift [tex]\( \phi = \frac{\pi}{12} \)[/tex], the equation is:
[tex]\[ y_2(x,t) = 7.00 \cos(3.00x - 2.50t + \frac{\pi}{12}) \]\\ At \( t = 2.00 \, \text{s} \) and \( x = 0.53 \, \text{m} \), we substitute these values into the equations for \( y_1 \) and \( y_2 \): \[ y_1(0.53,2.00) = 7.00 \cos(3.00 \cdot 0.53 - 2.50 \cdot 2.00) \] \[ y_1(0.53,2.00) = 7.00 \cos(1.59 - 5.00) \] \[ y_1(0.53,2.00) = 7.00 \cos(-3.41) \][/tex]
Since the cosine function is even, [tex]\( \cos(-3.41) = \cos(3.41) \)[/tex], we can write:
[tex]\[ y_1(0.53,2.00) = 7.00 \cos(3.41) \][/tex]
For the second wave:
[tex]\[ y_2(0.53,2.00) = 7.00 \cos(3.00 \cdot 0.53 - 2.50 \cdot 2.00 + \frac{\pi}{12}) \] \[ y_2(0.53,2.00) = 7.00 \cos(1.59 - 5.00 + \frac{\pi}{12}) \] \[ y_2(0.53,2.00) = 7.00 \cos(-3.41 + \frac{\pi}{12}) \] The resultant wave \( y \) is the sum of \( y_1 \) and \( y_2 \): \[ y(0.53,2.00) = y_1(0.53,2.00) + y_2(0.53,2.00) \] \[ y(0.53,2.00) = 7.00 \cos(3.41) + 7.00 \cos(-3.41 + \frac{\pi}{12}) \][/tex]
To find the numerical value, we calculate the cosines:
[tex]\[ y(0.53,2.00) = 7.00 \cos(3.41) + 7.00 \cos(-3.41 + \frac{\pi}{12}) \] \[ y(0.53,2.00) = 7.00 \cos(3.41) + 7.00 \cos(-3.41 + 0.2618) \] \[ y(0.53,2.00) = 7.00 \cos(3.41) + 7.00 \cos(-3.1482) \][/tex]
Using a calculator or trigonometric tables, we find:
[tex]\[ y(0.53,2.00) = 7.00 \cdot (-0.9956) + 7.00 \cdot (-0.9994) \] \[ y(0.53,2.00) = -6.9692 - 6.9958 \] \[ y(0.53,2.00) = -13.965 \][/tex]
Therefore, the height of the resultant wave at [tex]\( t = 2.00 \, \text{s} \) and \( x = 0.53 \, \text{m} \)[/tex] is:
[tex]\[ \boxed{-13.965 \, \text{cm}} \][/tex]
The negative sign indicates that the displacement is in the negative y-direction.
A parachutist of mass 39.4 kg jumps out of an airplane at a height of 1340 m and lands on the ground with a speed of 5.78 m/s. The acceleration of gravity is 9.8 m/s². How much energy was lost to air friction during this jump?
Answer:
Approximately 500kJ
Explanation:
According to the law of conservation of energy which states that energy can neither be created nor destroyed but can be converted from one form to another.
A parachutist jumping out of the airplane covering a particular height and under the influence of gravity will possess potential energy during fall. Since PE = mass × acceleration due to gravity × height
PE = 39.4×1340×9.8
PE = 501,562Joules
If the body lands on the ground with a speed of 5.78m/s, this means the body possesses kinetic energy at the point of landing. The kinetic energy on landing is 1/2mv²
KE = 1/2×39.4×5.78²
KE = 658.14Joules
The amount of energy lost due to friction will be PE-KE
= 501,562-658.14
= 500,903Joules approximately 500kJ
Final answer:
The solution involves calculating the parachutist's initial potential energy and final kinetic energy, then finding the difference to determine the energy lost to air friction during the jump.
Explanation:
The question asks how much energy was lost to air friction during the jump of a parachutist who has a mass of 39.4 kg, jumps from a height of 1340 m, and lands with a speed of 5.78 m/s. To find the energy lost to air friction, we first calculate the potential energy at the start and the kinetic energy at the end of the jump, then find the difference.
Initial potential energy (PE_initial) is given by mgh, where m is mass, g is acceleration due to gravity (9.8 m/s²), and h is height. Therefore, PE_initial = 39.4 kg * 9.8 m/s² * 1340 m.
Final kinetic energy (KE_final) is given by 1/2 mv², where m is mass and v is velocity at landing. Thus, KE_final = 1/2 * 39.4 kg * (5.78 m/s)².
The energy lost to air friction equals the initial potential energy minus the final kinetic energy. Subtracting KE_final from PE_initial gives the amount of energy lost to air resistance.
explain how stability and toxicity can work together to increase the effect of a chemical on the environment
Answer:
The concepts of basic chemistry and physics teach us that the more stable a compound would be, the more difficult it will be to break that compound. On the other hand, compounds which are not stable have weak binds and can be broken down easily.
For example, CFC's are stable compounds. It is difficult to break down these compounds and hence they exist in the environment without being broken down. They cause harmful effects on the ozone layer of the Earth.
Two mountain bikers are rolling down a hill. They both have the same mass, but one is taller and has more air drag (friction, etc.) than the shorter one. Which will reach the bottom first?
Answer:The Taller one
Explanation: Air drag(friction) also known as air resistance,the opposing force acting against the force of gravity, if a falling Object have a high air drag it will mean the object have a high air resistance. The higher the air resistance the higher the impact of the force of gravity on an object. The force of gravity will act to want to overcome the opposing force in order to bring the Object or the taller person down.
In a situation with no energy losses, an object sliding without rolling will reach the bottom of an incline first since all its energy goes into descending speed. For two mountain bikers with the same mass, the one with less air resistance will reach the bottom first.
The question concerns the concept of rolling motion versus sliding motion and its effect on the speed at which objects descend an incline. In an ideal scenario with no energy losses, an object that is rolling will have some of its energy in rotational kinetic energy in contrast to an object that is sliding which will only have translational kinetic energy. As a result, the object sliding without rolling, given the same initial conditions, will reach the bottom first due to having all of its energy contributing to its descent speed.
In the case of two mountain bikers with the same mass but differing in air resistance, the biker with less air drag will reach the bottom first. Air resistance acts as a form of friction that takes away some of the energy that could otherwise be converted into speed, thereby slowing down the biker.
Both Kepler’s laws and Newton’s laws tell us something about the motion of the planets, but there are fundamental differences between them. What are the differences? Select all of the true statements.
Kepler's laws describe the motion of planets based on empirical observations without explaining why they behave in such a way. Newton's laws, particularly the law of universal gravitation, provide the reasons for these behaviors, explaining them using cause-effect relationships.
The fundamental differences between Kepler's laws and Newton's laws pertain to their descriptions of planetary motion. Kepler's laws are based on the empirical observations and provide a descriptive perspective on the motion of planets around the Sun in an elliptical orbit. They consist of three essential laws: the Law of Orbits, Law of Areas, and Law of Periods.
Newton's laws, on the other hand, provide an explanatory perspective grounded in cause-effect relationships. These laws, especially the law of universal gravitation, explain why the planets follow an elliptical orbit as Kepler observed, based on the gravitational influence of the Sun.
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The differences between Kepler's laws and Newton's laws are that Kepler described planetary motion through empirical observations, while Newton explained it theoretically with universal laws applicable to all motions in the universe.
The true statements regarding the differences between Kepler's laws and Newton's laws are:
A. Kepler reported how planets moved, and Newton explained why.
Kepler's laws describe the observed motions of planets, while Newton's laws provide a theoretical explanation for these motions based on the law of universal gravitation.
B. Kepler defined laws based on one special case - the observed motions of planets in the Solar System, while Newton defined theoretical laws that describe the general behavior of all motions in the universe.
Kepler's laws were specific to planetary motion in the Solar System, while Newton's laws of motion and the law of universal gravitation are general laws that apply to all objects and motions in the universe.
D. Kepler's approach was empirical, while Newton's was theoretical.
Kepler's laws were derived from observations, making his approach empirical. In contrast, Newton's laws were based on theoretical principles, including the law of universal gravitation.
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The question is incomplete, the given complete question is:
"Both Kepler's laws and Newton's laws tell us something about the motion of the planets, but there are fundamental differences between them. What are the differences? Select all of the true statements. Choose one or more: O A. Kepler reported how planets moved, and Newton explained why. B. Kepler defined laws based on one special case-the observed motions of planets in the Solar System, while Newton defined theoretical laws that describe the general behavior of all motions in the universe O C. Kepler's discoveries built on the work of an earlier astronomer, while Newton's did not. OD. Kepler's approach was empirical, while Newton's was theoretical."
In a nuclear power plant, the temperature of the water in the reactor is above 100°C because of what?
Answer:
The temperature of the water increases because the nuclear reactor heats it producing steam
Explanation:
The nuclear power plants are usually defined as those thermal plants where the nuclear reactors are used in order to generate heat that eventually leads to the rotating of the turbines and produces electricity. Here the nuclear reactor heats the water, and it increases above a temperature of 100°C, where this heat energy plays a key role in the entire process. It is an efficient method as it does not lead to the emission of any green house gases that are harmful to the environment.
Technician A says dc circuits are not normally used in automotive circuits. Technician B says certain automotive computer sensors use ac. Who is right?
Answer:
A & B are correct
Explanation:
Certain automotive sensors like crank shaft position sensor uses a.c. while most automotive components require DC to work properly through the battery which is incorporated with an alternator to keep it charged through a diode while continual supply of DC is sustained.
Technician A is incorrect as DC circuits are commonly used in automobiles, and Technician B is partially correct since some automotive sensors can generate an AC signal.
Explanation:In the scenario presented, Technician A says that dc circuits are not normally used in automotive circuits, which is incorrect. DC circuits are indeed commonly used in automobiles for various functions such as starting the engine, powering lights, and running the dashboard instruments. Technician B says that certain automotive computer sensors use ac current. While most sensors in a car operate on DC current, there are some sensors, like variable reluctance sensors, that can generate an AC signal as they operate. Therefore, both statements are not entirely accurate; however, Technician B's statement has some validity regarding the existence of sensors that produce an AC signal within automotive applications.
Solar wind particles can be captured by the Earth's magnetosphere. When these particles spiral down along the magnetic field into the atmosphere, they are responsible for ?
Answer:
Solar wind particles can be captured by the Earth's magnetosphere. When these particles spiral down along the magnetic field into the atmosphere, they are responsible for?
Answer is Aurorae.
Explanation:
Solar Wind:
A planet's magnetic filed forms a shield to protect the surface of planet from energetic and charged particles coming from sun and other planets. The sun is continuously sending out charged particles , called Solar wind.
Magnetosphere of planet:
If a planet has a magnetic field then it will interact with the solar wind and deflect the charged particles in the wind. Due to which an elongated cavity in solar wind formed. This cavity is called magnetosphere of the planet.
When Solar wind particles run in a magnetic field , they are deflected and spiral down along magnetic field lines into the atmosphere.
Most of the solar wind particles deflected around the planet but a few particles manage to leak into the magnetic filed and become trapped in the magnetic field of the planet, to create Radiation Belts or Charged Particles Belts.
Variable Solar Wind can give some radiation belt particles with enough energy to spiral down along the magnetic filed into the atmosphere and Create Aurorae
Here Aurorae is an atmospheric phenomenon consisting of Bands, streamers of light, usually yellow , green or red , that move across the sky in the polar regions.
The charged particles from the solar wind spiral down along the Earth's magnetic field lines toward the poles, causing the atmospheric gases they collide with to emit light, resulting in auroras, commonly known as northern and southern lights.
Explanation:When charged particles, known as the solar wind, emitted from the Sun encounter the Earth's magnetosphere, they can get trapped and follow the magnetic field lines towards the Earth's poles. Upon interacting with atmospheric gases, these charged particles cause a natural light display in the sky known as the auroras. These spectacular light shows are called the aurora borealis, or northern lights, in the Northern Hemisphere, and the aurora australis, or southern lights, in the Southern Hemisphere.
These auroras are more than just beautiful; they represent the interaction between the Sun's energy and our planet's protective magnetic field. During periods of intense solar activity, the solar wind is stronger, leading to more significant and vibrant auroral displays. These geomagnetic storms caused by solar activity can also have consequential effects on power grids, satellite operations, and communication systems.
Two cars are traveling along a straight road. Car A maintains a constant speed of 95 km/h and car B maintains a constant speed of 121 km/h. At t = 0, car B is 41 km behind car A.
(a) How much farther will car A travel before car B overtakes it?km
(b) How much ahead of A will B be 30 s after it overtakes A?km
Car A will travel 150.1 km further before Car B overtakes it, and 30 seconds after Car B overtakes Car A, Car B will be 1.005 km ahead of Car A.
Explanation:This is a physics problem involving relative speed and time. We are trying to find out how much farther Car A will travel before Car B catches up, as well as Car B's lead distance 30 seconds after it overtakes Car A.
a) How much farther will car A travel before car B overtakes it?
First, we calculate the Relative Speed of the two cars: Relative Speed = Speed of B - Speed of A = 121 km/h - 95 km/h = 26 km/h. Then we find out how long it takes for Car B to close that 41 km gap at this relative speed: Time = Distance/Speed = 41km / 26km/h = 1.58 hours. Therefore, in this duration, Car A will travel Distance = Speed * Time = 95km/h * 1.58h = 150.1 km.
b) How much ahead of A will B be 30 s after it overtakes A?
Here we just calculate the distance Car B travels in 30 seconds (converted to hours). Distance = Speed * Time = 121km/h * (30s / 3600s/h) = 1.005 km.
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Car A will travel approximately 150.1 km before Car B overtakes it. Car B will be approximately 0.22 km ahead of Car A 30 seconds after overtaking. The solution involves calculating the time taken for Car B to close the initial gap and then the additional distance traveled in the subsequent 30 seconds.
Part (a)
To determine how much farther Car A will travel before Car B overtakes it, we can set up an equation based on the relative speeds of the two cars and the distance between them.
Let the time taken for Car B to overtake Car A be t hours.
In time t, Car A travels 95t km.
In time t, Car B travels 121t km.
At the point of overtaking, the distance covered by Car A plus the initial 41 km (since Car B started 41 km behind) will be equal to the distance covered by Car B:
95t + 41 = 121t
Solving for t:
121t - 95t = 41
26t = 41
t = 41/26 ≈ 1.58 hours
The distance Car A travels in this time is:
95 km/h × 1.58 hours ≈ 150.1 km
Part (b)
To find out how far ahead Car B will be 30 seconds after overtaking Car A:
Convert 30 seconds to hours:
30 seconds = 30/3600 hours = 1/120 hours.
In 1/120 hours :
Car A travels: 95 km/h × 1/120 hours ≈ 0.79 km.
Car B travels: 121 km/h × 1/120 hours ≈ 1.01 km.
The difference in distance traveled by Car B and Car A:
1.01 km - 0.79 km ≈ 0.22 km.
So, Car B will be approximately 0.22 km ahead of Car A 30 seconds after overtaking it.
A transformer has a 240 V primary and a 60 V secondary. With a 5 ohm load connected, what is the primary current?
The primary current for the given transformer is 3 A which is determined by considering the principle of a transformer, the turns ratio and the law of conservation of power, and calculating secondary current using Ohm's law.
Explanation:To find the primary current of a transformer, we need to first understand the principle of a transformer. A transformer operates on the electromagnetic induction principle, and the ratio of turns in the primary and secondary coil determines how much voltage is changed. In your transformer, the primary voltage is 240 V and the secondary voltage is 60 V. Therefore, the turns ratio is 240/60 which gives a value of 4.
Now, considering the law of conservation of power in an ideal transformer, the power output is equal to the power input. Power is given by the product of voltage and current.
So, primary voltage (V1) times primary current (I1) is equal to secondary voltage (V2) times secondary current (I2). The secondary current (I2) can be calculated using Ohm's law as V2/R, where R is the resistance of the load connected. Plugging in the given values we get
I2 = 60/5 = 12 A.
Now, substituting the known variables in the power conservation equation we get
240*I1 = 60*12.
Solving for I1 gives a value of 3A.
Therefore, the primary current for the given transformer is 3 A.
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A 0.500 kg pendulum bob passes through the lowest part of its path at a speed of 3.40 m/s.
(a) What is the tension in the pendulum cable at this point if the pendulum is 80.0 cm long? (in Newtons)
(b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? (in Degrees)
(c) What is the tension in the pendulum cable when the pendulum reaches its highest point? (in Newtons)
Final answer:
The tension in the pendulum at the lowest point is 12.1 N. Without additional information, we cannot determine the exact angle at which the pendulum reaches its highest point. The tension in the pendulum at the highest point is 4.9 N.
Explanation:
Tension in the Pendulum Cable
To determine the tension in the pendulum cable at the lowest point, we can use the formula for circular motion.
T = mg + m(v^2/r)
Where:
g is the acceleration due to gravity (9.8 m/s²)
r is the length of the pendulum
Substituting the values given, we have:
T = 0.500 kg * 9.8 m/s² + 0.500 kg * (3.40 m/s)^2 / 0.80 m
Now, we can calculate the tension:
T = 4.9 N + 7.225 N
T ≈ 12.1 N
The Angle at the Highest Point
To find the angle at the highest point, we use the principle of conservation of energy. The kinetic energy at the lowest point equals potential energy at the highest point.
mgh = 1/2 mv^2
Solving for h gives us h = v^2/(2g), and then use trigonometry to find the angle θ.
h = r(1 - cosθ)
Without sufficient information to calculate the speed of the pendulum, we must leave the answer as an open calculation.
Tension at the Highest Point
At the highest point, the tension is equal to the weight of the pendulum alone since it is momentarily at rest, and there is no centripetal force required.
T = mg
Thus, the tension at the highest point will be:
T = 0.500 kg * 9.8 m/s² = 4.9 N.
What frequency (in Hz) is received by a person watching an oncoming ambulance moving at 108 km/h and emitting a steady 900 Hz sound from its siren? The speed of sound on this day is 345 m/s.
Answer:985.71 Hz
Explanation:
Given
Speed of ambulance (source) [tex]v_s=108\ kmph\approx 30\ m/s[/tex]
Original Frequency of sound wave [tex]f=900\ Hz[/tex]
speed of sound waves [tex]v=345\ m/s[/tex]
According to Doppler apparent frequency will be different to original frequency when speed of source is moving w.r.t to observer.
[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex]
where f'=apparent frequency
f=original frequency
[tex]v_o[/tex]=observer speed
[tex]f'=900\times (\frac{345+0}{345-30})[/tex]
[tex]f'=900\times (\frac{345}{315})[/tex]
[tex]f'=985.71\ Hz[/tex]
A balloonist, riding in the basket of a hot air balloon that is rising vertically with a constant velocity of 11.9 m/s, releases a sandbag when the balloon is 43.4 m above the ground. Neglecting air resistance, what is the bag's speed when it hits the ground?
Answer:
v = 31.5 m/s
Explanation:
given,
Speed of air balloon, u = 11.9 m/s
height of balloon above the ground, h = 43.4 m
acceleration due to gravity = 9.8 m/s²
Speed of the bag when it hit the ground = ?
using equation of motion
[tex]v^2 = u^2 + 2 g h[/tex]
[tex]v^2 = 11.9^2 + 2\times 9.8\times 43.4[/tex]
[tex]v= \sqrt{992.25}[/tex]
v = 31.5 m/s
Speed of the bag before it hit the ground is equal to v = 31.5 m/s
Applying the physics of kinematics, specifically the equations of motion, the speed of the sandbag when it hits the ground can be calculated to be 29 m/s.
Explanation:The speed of the sandbag when it hits the ground can be calculated with the concepts of kinematics in physics, in particular, the equations of motion. Considering that the positive direction is upwards and the balloon is moving upward at a velocity of 11.9 m/s, the initial velocity (vi) of the sandbag is +11.9 m/s. The sandbag is then acted upon by gravity, which imparts a downward acceleration of -9.8 m/s2. The final speed (vf), when the sandbag hits the ground, can be calculated using the equation vf2 = vi2 + 2ad, where a is acceleration and d is the distance. Substituting the given values: vf2 = (11.9 m/s)2 + 2*(-9.8 m/s2)*(-43.4 m), we get vf ≈ 29 m/s, which is the speed of the sandbag when it hits the ground.
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