Answer:
a) 2.67 m/s
b) 0.82 J
Explanation:
Amplitude A = 25 cm = 0.25 m
The period of the motion is the inverse of the frequency
[tex]T = \frac{1}{f} = \frac{1}{1.7} = 0.588 s[/tex]
So the angular frequency
[tex]\omega = \frac{2\pi}{T} = \frac{2\pi}{0.588} = 10.68 rad/s[/tex]
The speed at the equilibrium point is the maximum speed, at
[tex]v = \omega A = 10.68 * 0.25 = 2.67 m/s[/tex]
The spring constant can be calculated using the following
[tex]\omega^2 = \frac{k}{m} = \frac{k}{0.23}[/tex]
[tex]k = 0.23\omega^2 = 0.23*10.68^2 = 26.24 N/m[/tex]
The total energy of the oscillation is
[tex]E = kA^2 / 2 = 26.24*0.25^2 / 2 = 0.82 J[/tex]
Radio wave radiation falls in the wavelength region of 10.0 to 1000 meters. What is the energy of radio wave radiation that has a wavelength of 784 m?
Radio waves have lower energy compared to other types of waves in the electromagnetic spectrum. The energy of a radio wave with a wavelength of 784 m can be calculated using the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency.
Explanation:Radio waves fall within the electromagnetic spectrum which consists of various types of waves ranging from gamma rays to radio waves. The energy of a wave is related to its wavelength and frequency through the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency. Since radio waves have a longer wavelength, they have a lower frequency and therefore lower energy compared to other types of waves in the spectrum.
To calculate the energy of a radio wave with a wavelength of 784 m, we can use the equation c = λf, where c is the speed of light. Rearranging the equation to solve for f, we get f = c/λ. Plugging in the given wavelength of 784 m and the speed of light which is approximately 3 x 10^8 m/s, we can calculate the frequency as 3 x 10^8 m/s / 784 m = 3.83 x 10^5 Hz. Substitute this frequency value into the equation E = hf to calculate the energy of the radio wave.
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The energy of radio wave radiation for a 784-meter wavelength is approximately 2.53 x 10^-34 joules, calculated using Planck's equation for energy.
The energy E of radio wave radiation with a particular wavelength λ can be calculated using the equation:
E = hc/λ
Where:
E is the energy in Joules (J)
h is Planck's constant (6.626 x 10-34 J·s)
c is the speed of light in vacuum (about 3 x 108 m/s)
λ is the wavelength in meters (m)
For a radio wave with a wavelength of 784 meters, we use these values to compute the energy:
E = (6.626 x 10-34 J·s) * (3 x 108 m/s) / 784 m
Calculating using the numerical values:
E = (6.626 x 10-34) * (3 x 108) / 784
E ≈ 2.53 x 10-34 J
Therefore, the energy of radio wave radiation with a wavelength of 784 meters is approximately 2.53 x 10-34 joules.
A capacitor is constructed with two parallel metal plates each with an area of 0.52 m2 and separated byd = 0.80 cm. The two plates are connected to a 5.0-volt battery. The current continues until a charge of magnitude Q accumulates on each of the oppositely charged plates.
Find the electric field in the region between the two plates.
V/m
Find the charge Q.
C
Find the capacitance of the parallel plates.
? 10?6 F
(i) 625 V/m
(ii) 2876.25 x 10⁻¹² C
(iii) 0.000575.25 x 10⁻⁶ F
Explanation:(i) The electric field (E) between the plates of a parallel plate capacitor is related to the potential difference (V) between the plates and the distance (d) of separation between the plates as follows;
E = V / d ----------------(i)
From the question;
V = 5.0V
d = 0.80cm = 0.008m
Substitute these values into equation (i) as follows;
E = 5.0 / 0.008
Solve for E;
E = 625 V/m
Therefore, the electric field in the region between the two plates is 625 V/m.
(ii) To make things easier, let's calculate the capacitance of the parallel plates first.
The capacitance (C) of a parallel plate capacitor is given as;
C = A x ε₀ / d --------------------------(ii)
Where;
A = Area of either of the plates of the capacitor = 0.52m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹² F/m
d = distance between the plates = 0.8cm = 0.008m
Substitute these values into equation (ii) as follows;
C = 0.52 x 8.85 x 10⁻¹² / 0.008
Solve for C;
C = 575.25 x 10⁻¹² F
The capacitance (C) is related to potential difference (V) and charge (Q) on the plates as follows;
Q = C x V -------------------------(iii)
Where;
C = 575.25 x 10⁻¹² F
V = 5.0V
Substitute these values into equation (iii)
Q = 575.25 x 10⁻¹² x 5
Q = 2876.25 x 10⁻¹² C
Therefore, the charge on the plates is 2876.25 x 10⁻¹² C
(iii) The capacitance (C) of the parallel plates has been calculated in (ii) above.
Its value is 575.25 x 10⁻¹² F = 0.000575.25 x 10⁻⁶F
The electric field in the region between two plates is 625 V/m. The capacitance of the plates is 0.0572 nanoFarads and the accumulated charge Q on each plate is 2.86 x 10^-10 Coulombs.
Explanation:The electric field between two plates can be calculated using the formula E = V/d, where V represents voltage and d represents the distance between plates. So, the electric field E would be 5.0 V / 0.008 m = 625 N/C or 625 V/m. The charge Q on the plates can be determined using the formula Q = CV, where C is the capacitance and V is the voltage. To find the charge, we first need to calculate the capacitance. The capacitance of the parallel plates can be calculated using the formula C = ε0*(A/d), where ε0 is the permittivity of free space (8.85 x 10^-12 F/m), A is the area and d is the distance between the plates. The capacitance then would be 8.85 x 10^-12 F/m * 0.52 m^2 / 0.008 m = 5.72 x 10^-11 F or 0.0572 nF (nanoFarads). So, after finding the capacitance, we can now calculate the charge Q, which is Q = CV = 5.72 x 10^-11 F * 5.0 V = 2.86 x 10^-10 C (Coulombs).
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In the afternoon, the decibel level of a busy freeway is 80 dB with 100 cars passing a given point every minute. Late at night, the traffic flow is only 5 cars per minute. What is the late-night decibel level?
Answer:
67 dB
Explanation:
given,
Sound in afternoon = 80 dB
Intensity of car,I₀ = 100 cars/ minute
Sound in the night = ?
Intensity of car,I = 5 car/minutes
using formula for sound calculation
[tex]\Delta \beta = 10 log(\dfrac{I}{I_0})[/tex]
[tex]\Delta \beta = 10 log(\dfrac{5}{100})[/tex]
[tex]\Delta \beta = 10 log(\dfrac{1}{20})[/tex]
[tex]\Delta \beta = 10 log(0.05)[/tex]
[tex]\Delta \beta = 10\times -1.30[/tex]
[tex]\Delta \beta = -13\ dB[/tex]
The late night decibel is equal to 80 dB - 13 dB = 67 dB
On the moon, the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is "weighed" with a beam balance on the surface of the moon.
a) What is the expected reading?
b) If this mass is weighed with a spring scale that reads correctly for standard gravity on earth, what is the reading?
Answer:
A. 8.175 N
B. 49.05 N
Explanation:
A.
Acceleration due to gravity (moon) = 1/6 * (acceleration due to gravity (earth)
Acceleration due to gravity (earth) = 9.81 m/s2
Acceleration due to gravity (moon) = 9.81/6
= 1.635 m/s
Weight. Fm = acceleration due to gravity * mass
= 1.635 * 5
= 8.175 N
B. Acceleration due to gravity (earth) * mass = Fe
= 9.81 * 5
= 49.05 N
List three advantages of reflecting telescopes over refracting telescopes.
Answer:
Three advantages of Reflecting Telescope over refracting Telescope:
1. there is no chromatic aberration in the Reflecting Telescope being mirror as an objective, while Refracting Telescope can suffer chromatic aberration being using lenses.
2. image of Reflecting Telescope is brighter due to large, polished curved mirrors than Refracting Telescope
3. Reflecting Telescope is compact and portable in size than Refracting Telescope.
Final answer:
Reflecting telescopes have three advantages over refracting telescopes: they gather more light, do not suffer from chromatic aberration, and are easier and less expensive to manufacture.
Explanation:
There are three advantages of reflecting telescopes over refracting telescopes:
Light gathering: Reflecting telescopes can gather more light than refracting telescopes because the mirror in a reflecting telescope is larger in diameter than the lens in a refracting telescope. This allows reflecting telescopes to have a greater ability to capture faint objects in the sky.No chromatic aberration: Reflecting telescopes do not suffer from chromatic aberration, which is a distortion of colors caused by different wavelengths of light focusing at different points. This is because reflecting telescopes use mirrors instead of lenses to focus light, eliminating the issue of chromatic aberration.Ease and cost of manufacturing: Reflecting telescopes are easier and less expensive to manufacture than refracting telescopes. Only the front surface of a mirror needs to be accurately polished, whereas lenses in a refracting telescope require both sides to be polished to great accuracy. Reflecting telescopes also do not require high-quality glass throughout like refracting telescopes.Electrons in a particle beam each have a kinetic energy of 4.0 × 10−17 J. What is the magnitude of the electric field that will stop these electrons in a distance of 0.3 m? (e = 1.6 × 10−19 C) Group of answer choices
Explanation:
Relation between work and change in kinetic energy is as follows.
[tex]W_{net} = \Delta K[/tex]
Also, [tex]\Delta K = K_{initial} - K_{final}[/tex]
= [tex](0 - 4.0 \times 10^{-17})[/tex] J
= [tex]-4.0 \times 10^{-17}[/tex] J
Let us assume that electric force on the electron has a magnitude F. The electron moves at a distance of 0.3 m opposite to the direction of the force so that work done is as follows.
w = -Fd
[tex]-4.0 \times 10^{-17} J = -F \times 0.3 m[/tex]
F = [tex]1.33 \times 10^{-16}[/tex]
Therefore, relation between electric field and force is as follows.
E = [tex]\frac{F}{q}[/tex]
= [tex]\frac{1.33 \times 10^{-16}}{1.60 \times 10^{-19} C}[/tex]
= [tex]0.831 \times 10^{3}[/tex] C
Thus, we can conclude that magnitude of the electric field that will stop these electrons in a distance of 0.3 m is [tex]0.831 \times 10^{3}[/tex] C.
An elevator in a tall building is allowed to reach a maximum speed of 3.3 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 3.4 m if the elevator has a mass of 1320 kg including occupants?
To calculate the tension in the cable required to stop the elevator, we multiply the mass of the elevator by its acceleration. The tension in the cable is -4356 N, indicating it acts in the opposite direction of the weight of the elevator.
Explanation:To calculate the tension in the cable required to stop the elevator over a distance of 3.4 m, we need to consider the force required to decelerate the elevator from its maximum speed of 3.3 m/s to a stop. The tension in the cable must equal the force needed to stop the elevator, which is equal to the mass of the elevator multiplied by its acceleration. The mass of the elevator, including occupants, is given as 1320 kg. Since the elevator is going down, its acceleration will be negative. Therefore, the tension in the cable can be calculated using the formula:
Tension = mass * acceleration = 1320 kg * (-3.3 m/s^2) = -4356 N
Therefore, the tension in the cable to stop the elevator over a distance of 3.4 m is -4356 N. The negative sign indicates that the tension is acting in the opposite direction of the weight of the elevator.
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You are standing in a boat. Which of the following strategies will make the boat start moving? Check all that apply.
a. Pushing its mast
b. Throwing some cargo out of the boat
c. Pushing the front of the boat
d. Pushing another passenger
Answer:
b. Throwing some cargo out of the boat
Explanation:
Using the Newton's third law of motion which states that every action has an equal and opposite reaction.
So when we are on the boat and we throw some mass in a direction away from the boat out of it then we are imparting the force to the floating boat by the law of conservation of momentum as well.
The relation can be mathematically expressed as:
[tex]m_c.v_c=m_b.v_b[/tex]
where:
[tex]m_c=[/tex] mass of the cargo
[tex]v_c=[/tex] velocity of throwing the cargo
[tex]m_b=[/tex] mass of the whole boat including all that floats on it
[tex]v_b=[/tex] velocity of the boat system
Throwing some cargo out of the boat will make the boat start moving.
Newton's third law of motion;This law states that action and reaction are equal and opposite.
The force apply in pushing the mast, or another passenger in the boat will be equal to the force they will push back at you. The two forces will cancel out and the boat will remain stationary.Principle of conservation of linear momentum;The total momentum of an isolated system is always conserved.
Throwing some cargo out of the boat will change the velocity of the cargo initially at rest and to conserve the linear momentum, the velocity of the boat will change as well.
Thus, throwing some cargo out of the boat will make the boat start moving.
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At a certain distance from a point charge, the potential and electric field magnitude due to that charge are 4.98 V and 12.0 V/m, respectively. (Take the potential to be zero at infinity.)1.What is the distance to the point charge? (d= ? m)2.What is the magnitude of the charge? (q= ? c)
Answer:
1. d = 0.415 m.
2. Q = 2.285 x 10^{-10} C.
Explanation:
The electric field and potential can be found by the following equations:
[tex]E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\\V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}[/tex]
Applying these equations to the given variables yields
[tex]E = 12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}\\V = 4.98 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d}[/tex]
Divide the first line to the second line:
[tex]\frac{12}{4.98} = \frac{ \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}}{\frac{1}{4\pi\epsilon_0}\frac{Q}{d}}\\\frac{12}{4.98} = \frac{1}{d}\\d = 0.415~m[/tex]
Using this distance in either of the equations give the magnitude of the charge.
[tex]12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{(0.415)^2}\\12 = \frac{1}{4\pi (8.8\times 10^{-12})}\frac{Q}{(0.415)^2}\\Q = 2.285 \times 10^{-10}~C[/tex]
Final answer:
The distance to the point charge is 0.415 meters. The magnitude of the charge is 231 picocoulombs (pC).
Explanation:
To calculate the distance to the point charge when the electric potential is 4.98 V and electric field magnitude is 12.0 V/m, we can use the relationship between electric field (E) and electric potential (V), which is E = -dV/dr. Rearranging the equation to solve for the distance (r), we get r = V/E = 4.98 V / 12.0 V/m = 0.415 m.
To find the magnitude of the charge (q), we can use the electric potential formula for a point charge: V = k*q/r, where k is the Coulomb's constant (8.99 x [tex]10^9 Nm^2/C^2[/tex]). Rearranging the formula and solving for q, we get q = V*r/k = (4.98 V * 0.415 m) / (8.99 x [tex]10^9 Nm^2/C^2[/tex]) = 2.31 x [tex]10^-^1^0[/tex] C or 231 pC.
If a fixed length simple pendulum is found to have three times the period on an unknown planet’s surface (compared to Earth), what is the acceleration due to gravity on that planet? Show your work.
Answer:
g/9
Explanation:
length of the pendulum = L
time period on the earth = T
Time period on the planet = 3T
Let the acceleration due to gravity on the earth is g and on the planet is g'.
Use the formula for the time period of a simple pendulum for the time period on earth
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex] .... (1)
Time period on the surface of planet is
[tex]3T=2\pi \sqrt{\frac{L}{g'}}[/tex] .... (2)
Divide equation (2) by equation (1)
[tex]\frac{3T}{T}= \sqrt{\frac{g}{g'}}[/tex]
g' = g/9
Thus, the acceleration due to gravity on the planet is g /9
A pulse is transmitted down a long string made of two pieces of different materials. If the wavelength of the pulse received at the end is longer than at the beginning, this implies that the speed of the pulse in the second part of the string is
Options
1. the same as in the first.
2.greater than in the first
3.less than in the first.
4.Unable to determine
Answer
The answer is 2. Greater than in the first
Explanation
The speed of a wave v is related to its wavelength λ by the formulav=f λ, where f is the frequency of the wave. The frequency will not change when the wave passes into a second medium, so
λ2>λ1
Fλ1>fλ2
Since f>0
And V2>v1
According to the Guinness Book of World Records (1999) the highest rotary speed ever attained was 2010 m/s (4500 mph) The rotating rod was 15.3 cm long. Assume that the speed quoted is that of the end of the rod. a. What is the centripetal acceleration of the end of the rod? (2.64 x 107 m/s2) b. If you were to attach a 1.0 g object to the end of the rod, what force would be needed to hold it on the rod?
Answer:
a. 2.645 * 10^7 m/s^2
b. 2.645 * 10^4 N
Explanation:
Parameters given:
Velocity of rod = 2010m/s
Length of rod = 15.3cm = 0.153m
Mass of object placed at the end of the rod = 1g = 0.001kg
a. Centripetal acceleration is given as:
a = (v*v)/r
Where v = velocity
r = radius of curvature.
The radius of curvature in this case is equal to the length of the rod, since the rod makes the circular path of the motion.
Hence, centripetal acceleration at the end of the rod:
a = (2010*2010)/(0.153)
a = 26432156.86 m/s^2 = 2.64 * 10^7 m/s
b. The force needed to hold the object at the end of the rod is equal to the centripetal force at the end of the rod. Centripetal force is given as:
F = ma = (m*v*v)/r
Where a = centripetal acceleration
F = 0.001 * 2.64 * 10^7
F = 2.64 * 10^4N
The Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].
Given here:
Velocity of rod = 2010 m/s
Length of rod = 15.3cm = 0.153 m
Mass of object placed at the end of the rod = 1g = 0.001 kg
(A). Centripetal acceleration can be calculated by
[tex]\bold {a =\dfrac { v^2}{r}}[/tex]
Where
v = velocity
r = radius of curvature.
Put the values,
[tex]\bold {a = \dfrac {(2010^2)}{(0.153)}}\\\\\bold {a = 26432156.86\ m/s^2 } \\\\\bold {a = 2.64 x 10^7\ m/s}[/tex]
(B). Centripetal force can be calculated by
[tex]\bold {F = ma = \dfrac {(m v^2)}{r}}[/tex]
Where,
a = centripetal acceleration
Thus the force
[tex]\bold {F = 0.001 \times 2.64x10^7}\\\\\bold {F = 2.64x 10^4\ N}[/tex]
Therefore, the Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].
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A large box of mass M is moving on a horizontal surface at speed v0. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.
Part A: Find an expression for the shortest distance dminin which the large box can stop without the small box slipping.
Part B: A pickup truck with a steel bed is carrying a steel file cabinet. If the truck's speed is 10 m/s , what is the shortest distance in which it can stop without the file cabinet sliding? Assume that μs=0.80.
Answer:
Part A:
[tex]d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]
Part B:
[tex]d_{min} = 6.37~m[/tex]
Explanation:
Part A:
We should determine the free-body diagram of the small box.
For the first box, the only force exerted to the box is the static friction force in the direction of the motion.
(The direction of the static friction is always confusing to the students. The wrong idea is that the static friction is in the opposite direction with the motion. However, if you look at the Newton's Second Law, it states that the net force acting on an object is equal to the mass times acceleration. And in this case acceleration of the total system is equal to that of the small box, since it sits on the larger box.)
We can use the equations of kinematics to find the minimum distance to stop without the small box slipping.
[tex]v^2 = v_0^2 + 2a(\Delta x)\\0 = v_0^2 + 2ad_{min}[/tex]
The acceleration can be found by Newton's Second Law:
[tex]F = ma\\mg\mu_s = m(-a)\\a = -g\mu_s[/tex]
The negative sign comes from the fact that in order for the boxes to stop they have to apply a negative acceleration.
Now, we can combine the two equations to find the distance x:
[tex]0 = v_0^2 + 2ad_{min} = v_0^2 + 2(-g\mu_s)d_{min}\\d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]
Part B:
We can apply the above formula to the truck and file cabinet.
[tex]d_{min} = \frac{v_0^2}{2g\mu_s} = \frac{10^2}{2(9.8)(0.80)} = 6.37~m[/tex]
The expression for the shortest distance which the large box can stop without the small box slipping is [tex]d_{min} = \frac{v_0^2 }{2\mu_s g}[/tex]
The shortest distance in which the pickup can stop without the file cabinet sliding is 6.38 m.
The given parameters;
mass of the bigger box, = Mspeed of the bigger box, = v0mass of the small box, = mcoefficient of static friction, = μscoefficient of kinetic friction, = μkThe expression for the shortest distance which the large box can stop without the small box slipping is calculated as follows;
Apply work-energy theorem;
[tex]\mu_s F \times d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s (Mg)d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s gd_{min} = \frac{v_0^2}{2} \\\\d_{min} = \frac{v_0^2}{2\mu_s g}\\\\[/tex]
At the given speed and coefficient of static friction, the shortest distance in which the pickup can stop without the file cabinet sliding is calculated as;
[tex]d_{min} = \frac{v^2}{2\mu_s g} = \frac{10^2 }{2\times 0.8 \times 9.8} = 6.38 \ m[/tex]
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A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels upward and then downward to the ground at the base of the building. Let +y be upward, and neglect air resistance.For the rock's motion from the roof to the ground, what is the vertical component vav−yvav−y of its average velocity?
Express your answer in terms of acceleration due to gravity ggg, and the variables v0v0v_{0} and HHH.
The average vertical velocity of the rock thrown upwards from the height H to the ground is determined by the total vertical displacement divided by the total time, considering the downward acceleration due to gravity is negative.
Explanation:The vertical component v_{av-y} of the average velocity of a small rock thrown straight up from the edge of a roof of height H to the ground can be found by using the kinematic equations of motion under constant acceleration. Since the acceleration due to gravity (g) is acting downward, it is represented as negative in the equations. Considering the upwards direction as positive and the downward acceleration as negative ensures that the final answer for v_{av-y} takes into account the total displacement, which includes both the upward and the downward path of the rock. The average velocity is thus the total displacement divided by the total time taken. The rock initially travels upwards to a maximum height before falling to the ground, completing its motion.
How much heat is absorbed by a 28g iron skillet when its temperature rises from 10oC to 27oC?
_____ ___
Answer units
Amount of heat absorbed: 214 J
Explanation:
When an object absorbs heat, its temperature increases according to the equation
[tex]Q=mC\Delta T[/tex]
where
Q is the heat absorbed
m is the mass of the object
C is the specific heat capacity of the material
[tex]\Delta T[/tex] is the change in temperature
For the iron skillet in this problem:
m = 28 g = 0.028 kg is the mass
[tex]C=450 J/kg^{\circ}C[/tex] is the iron specific heat capacity
[tex]\Delta T = 27-10=17^{\circ}C[/tex] is the increase in temperature
Solving for Q, we find the amount of heat absorbed:
[tex]Q=(0.028)(450)(17)=214 J[/tex]
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As you drive away from a radio transmitter, the radio signal you receive from the station is shifted to longer wavelengths. (T/F)
The statement is true due to the Doppler Effect. As you move away from the radio transmitter, the radio signal you receive appears to have been shifted to longer wavelengths due to the changes in the observer-source distance.
Explanation:The statement, 'As you drive away from a radio transmitter, the radio signal you receive from the station is shifted to longer wavelengths,' is True. This is due to a phenomenon known as the Doppler Effect.
The Doppler Effect explains that the observed wavelength of electromagnetic radiation is longer (red shift) when the source moves away from the observer. This means the wavelength of the radiation from the radio station would appear to increase (shift to a longer wavelength) as you drive away from it.
To maintain that same energy level required for transmission, the radio station emits waves at a higher frequency (shorter wavelengths). But, as you move away, the radio waves appear to have a lower frequency (longer wavelengths). This happens because the waves get 'stretched out' as the distance between you (the observer) and the radio station (the source) increases.
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They are then fixed at positions that are 4.30 x 10-11 m apart. What is EPEfinal - EPEinitial, which is the change in the electric potential energy?
Answer:
The change in the electric potential energy is [tex]-3.2\times10^{-16}\ J[/tex]
Explanation:
Given that,
Distance [tex]d=4.30\times10^{-11}\ m[/tex]
suppose, Two particles with charges +6 e and -10 e are initially very far apart
We need to calculate the change in the electric potential energy
Using formula of energy
[tex]\text{electric potential energy}=\text{final electric potential energy-initial electric potential energy}[/tex]
[tex]EPE=EPE_{f}-EPE_{i}[/tex]
Here, initial electric potential energy= 0
final electric potential energy [tex]EPE_{f}=\dfrac{kq_{1}q_{2}}{r_{2}^2}[/tex]
Put the value into the formula
[tex]EPE=\dfrac{kq_{1}q_{2}}{r_{2}^2}+0[/tex]
Put the value into the formula
[tex]EPE=\dfrac{9\times10^{9}\times6\times1.6\times10^{-19}\times(-10\times1.6\times10^{-19})}{(4.30\times10^{-11})}[/tex]
[tex]EPE=-3.2\times10^{-16}\ J[/tex]
Hence, The change in the electric potential energy is [tex]-3.2\times10^{-16}\ J[/tex]
This is an incomplete question, here is a complete question.
Two particles with charges +6 e⁻ and -10 e⁻ are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 4.30 × 10⁻¹¹ m apart. What is EPE(final) - EPE(initial), which is the change in the electric potential energy?
Answer : The change in the electric potential energy is, [tex]-2.92\times 10^{-6}J[/tex]
Explanation : Given,
Formula used for electric potential energy of the two charges when they are separated is:
[tex]EPE=\frac{1}{4\pi \epsilon_0}\times {\frac{q_1\times q_2}{r^2}[/tex]
[tex]EPE=\frac{k\times q_1\times q_2}{r^2}[/tex]
where,
EPE = electric potential energy
k = [tex]\frac{1}{4\pi \epsilon_0}=8.99\times 10^9[/tex]
[tex]q_1[/tex] = charge on 1st particle = +6 e⁻ = [tex]6\times 10^{-19}C[/tex]
[tex]q_2[/tex] = charge on 2nd particle = -10 e⁻ = [tex]-10\times 10^{-19}C[/tex]
r = distance between two charges = [tex]4.30\times 10^{-11}m[/tex]
Now put all the given values in the above formula, we get:
[tex]EPE=\frac{(8.99\times 10^9)\times (6\times 10^{-19})\times (-10\times 10^{-19})}{(4.30\times 10^{-11})^2}[/tex]
[tex]EPE=-2.92\times 10^{-6}J[/tex]
Initially EPE = 0 J
Thus, [tex]EPE_{final}-EPE_{initial}=-2.92\times 10^{-6}J[/tex]
The positive sign indicate the attractive force and negative sign indicate the repulsive force.
Thus, the change in the electric potential energy is, [tex]-2.92\times 10^{-6}J[/tex]
An elevator is moving upward at a constant speed of 2.50 m/s. A bolt in the elevator ceiling 3.00 m above the elevator floor works loose and falls. (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor (b) according to an observer in the elevator? (c) According to an observer standing on one of the floor landings of the building? (d) According to the observer in part (c), what distance did the bolt travel between the ceiling and the floor of the elevator?
To answer the question, we use principles from mechanics, involving the concepts of free fall and relative motion. The time taken by the bolt to hit the floor is around 0.78s and has different velocities to an observer inside and outside the elevator. According to a ground observer, the bolt travels a total distance of 4.95m.
Explanation:To answer these questions, we need to involve the principles of physics, specifically mechanics dealing with motion. Firstly, since the bolt was initially at rest, and it falls under the influence of gravity, we use the formula for time in free fall t = √(2h/g), where h is the initial height (3.00 m) and g is the acceleration due to gravity (9.81 m/s²).
For (b), for an observer in the elevator, the bolt appears to fall straight down, so its velocity just as it hits the floor will be v = gt = 9.81*0.78 = 7.65 m/s.
For (c), for an observer on the floor landing, the elevator is moving upwards, thus the velocity of the bolt relative to the observer on the ground would be the sum of the falling velocity and the velocity of the elevator, so v = gt + elevator speed = 7.65 + 2.50 = 10.15 m/s.
For (d), the bolt would have traveled the height of the fall plus the distance the elevator traveled during the fall according to the ground observer. The distance the elevator moves is d = elevator speed * t = 2.50*0.78 = 1.95 m, therefore, the total distance the bolt travels is 3.00 m + 1.95 m = 4.95 m.
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One possible remnant of a supernova, called a neutron star, can have the density of a nucleus, while being the size of a small city. What would be the radius, in kilometers, of a neutron star with a mass 10 times that of the Sun? The radius of the Sun is 7 × 108 m and its mass is 1.99 × 1030 kg.
The radius of a neutron star with a mass 10 times that of the Sun is approximately 29.62 km.
Explanation:To calculate the radius of a neutron star with a mass 10 times that of the Sun, we can use the equation for the Schwarzschild radius:
R = 2GM / c^2
Where R is the radius, G is the gravitational constant, M is the mass, and c is the speed of light. Plugging in the values, we get:
R = 2 * 6.67x10^-11 (m^3/kg/s^2) * (10 * 1.99x10^30 kg) / (3x10^8 m/s)^2
R ≈ 29.62 km
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Equipotential surfaces are to be drawn 100 V apart near a very large uniformly charged metal plate carrying a surface charge density σ = 0.75 μC/m2. How far apart (in space) are the equipotential surfaces?
Electric field due to uniformly charged metal plate is given by,
[tex]E = \frac{\sigma}{(2\epsilon_0)}[/tex]
Here,
[tex]\sigma[/tex] = Charge density
[tex]\epsilon_0 =[/tex] Vacuum Permittivity
Our values are,
[tex]\sigma = 0.75 muC/m^2 = 0.75*10^-6 C/m^2[/tex]
[tex]\epsilon_0 = 8.85*10^-12 F\cdot m^{-1}[/tex]
Replacing we have,
[tex]E = \frac{(0.75*10^-6)}{(2*8.85*10^-12)}[/tex]
[tex]F = 42372.88N/C[/tex]
Now we have the relation where energy is equal to the change of the potential in a certain distance, then
[tex]E = \frac{V}{d}[/tex]
Rearranging for the distance
[tex]d = \frac{V}{E}[/tex]
[tex]d = \frac{100}{42372.88}[/tex]
[tex]d = 0.00236m[/tex]
[tex]d = 2.36mm[/tex]
Therefore the distance is 2.36mm
A 3" diameter germanium wafer that is 0.020" thick at 300K has 2.272*10^17 Arsenic atoms added to it. What is the resistivity of the wafer? Germanium has 4.42*10^22 atoms/cc, electron and hole mobilities are 3900 and 1900 cm^2/(V*s). What is the resistivity of the Ge in ohm*microns? From other similar questions I see they are not tking into consideration the volume of the wafer plus the unit conversion.
Explanation:
Formula to calculate resistivity is as follows.
[tex]\rho = \frac{1}{qN \mu}[/tex]
= [tex]\frac{1}{(1.6 \times 10^{-19}) \times 2.72 \times 10^{17} \times 3900}[/tex] ohm/cm
= [tex]5.89 \times 10^{-3} ohm/cm[/tex]
As germanium is an intrinsic conductor. Hence, resistivity of Ge is as follows.
[tex]\rho_{1} = \frac{1}{2qN_{o}\sqrt{\mu_{e}\mu_{r}}}[/tex]
= [tex]\frac{1}{2q(N_{A})^{\frac{1}{2}}\sqrt{\mu_{e}\mu_{r}}}[/tex]
= [tex]\frac{1}{2 \times (1.6 \times 10^{-19}) \times (\sqrt{4.42 \times 10^{22}})\sqrt{(3900)(1900)}}[/tex] ohm/cm
= 0.546 [tex]ohm (\mu m)^{-1}[/tex]
Thus, we can conclude that resistivity of the Ge is 0.546 [tex]ohm (\mu m)^{-1}[/tex].
A uniformly charged disk of radius 35.0 cm carries a charge density of
9.00×10^−3 C/m^2. Calculate the electric field on the axis of the disk at the following distances from the center of the disk.
a. 5.00 cm
b. 10.0 cm
c. 50.0 cm
d. 200 cm
The electric field of a uniformly charged disk at different points on its axis was computed using the formula for the electric field due to a charged disk. a. 20.16 N/C b. 17.7 N/C c. 3.57 N/C d. 0.225 N/C
Explanation:The student is asking for the calculation of the electric field at different points on the axis of a uniformly charged disk with a known charge density.
The formula to find the electric field E due to a charged disk along its axis at distance x from the center is given by E =((σ/2ε0) * (1 - (x/(sqrt(x^2 + r^2)))), where r is the radius of the disk and σ is the charge density. The constant ε0 is the permittivity of free space, and its value is approximately 8.85 x 10^-12 C^2/N·m^2.
a. For x=5.00 cm, E ~= 20.16 N/C
b. For x=10.0 cm, E ~= 17.7 N/C
c. For x=50.0 cm, E ~= 3.57 N/C
d. For x=200 cm, E ~= 0.225 N/C
These results reflect the fact that the strength of the electric field decreases as one moves farther away from the center of the disk.
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The electric fields at the respective distances are:
[tex]a. \( E(5.00 \, \text{cm}) = 4.369 \times 10^8 \, \text{N}/\text{C} \) b. \( E(10.0 \, \text{cm}) = 3.695 \times 10^8 \, \text{N}/\text{C} \) c. \( E(50.0 \, \text{cm}) = 9.187 \times 10^7 \, \text{N}/\text{C} \) d. \( E(200 \, \text{cm}) = 7.728 \times 10^6 \, \text{N}/\text{C} \)[/tex]
To calculate the electric field on the axis of a uniformly charged disk at a distance z from the center, we can use the formula derived from Gauss's law for a flat surface:
[tex]\[ E(z) = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + R^2}} \right) \][/tex]
where E(z) is the electric field at a distance z from the center of the disk, [tex]\( \sigma \)[/tex] is the surface charge density, [tex]\( \epsilon_0 \)[/tex] is the vacuum permittivity [tex](\( 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \))[/tex], and R is the radius of the disk.
Given:
- Radius of the disk, [tex]\( R = 35.0 \, \text{cm} = 0.35 \, \text{m} \)[/tex]
- Charge density, [tex]\( \sigma = 9.00 \times 10^{-3} \, \text{C}/\text{m}^2 \)[/tex]
- Vacuum permittivity,[tex]\( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \)[/tex]
Now, let's calculate the electric field at the given distances:
a. At z = 5.00[tex]\text{cm} = 0.05 \, \text{m} \)[/tex]:
[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.05}{\sqrt{0.05^2 + 0.35^2}} \right) \] \[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.05}{\sqrt{0.001225 + 0.1225}} \right) \][/tex]
[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.05}{0.3555} \right) \][/tex]
[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.1406 \right) \] \[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.8594 \] \[ E(0.05) = \frac{7.7346 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.05) = 4.369 \times 10^8 \, \text{N}/\text{C} \][/tex]
b. At z = 10.0[tex]\text{cm} = 0.1 \, \text{m} \)[/tex]:
[tex]\[ E(0.1) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.1}{\sqrt{0.1^2 + 0.35^2}} \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.1}{\sqrt{0.01 + 0.1225}} \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.1}{0.3674} \right) \][/tex]
[tex]\[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.2722 \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.7278 \] \[ E(0.1) = \frac{6.5402 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.1) = 3.695 \times 10^8 \, \text{N}/\text{C} \][/tex]
c. At z = 50.0[tex]\text{cm} = 0.5 \, \text{m} \)[/tex]:
[tex]\[ E(0.5) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.5}{\sqrt{0.5^2 + 0.35^2}} \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.5}{\sqrt{0.25 + 0.1225}} \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.5}{0.6104} \right) \][/tex]
[tex]\[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.8193 \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.1807 \] \[ E(0.5) = \frac{1.6263 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.5) = 9.187 \times 10^7 \, \text{N}/\text{C} \][/tex]
d. At z = 200 [tex]\text{cm} = 2.0 \, \text{m} \)[/tex]:
[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{2.0}{\sqrt{2.0^2 + 0.35^2}} \right) \] \[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{2.0}{\sqrt{4 + 0.1225}} \right) \][/tex]
[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{2.0}{2.0317} \right) \] \[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.9848 \right) \][/tex]
[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.0152 \] \[ E(2.0) = \frac{1.368 \times 10^{-4}}{17.7 \times 10^{-12}} \] \[ E(2.0) = 7.728 \times 10^6 \, \text{N}/\text{C} \][/tex]
A molecule is Select one: a. a carrier of one or more extra neutrons. b. a combination of two or more atoms. c. less stable than its constituent atoms separated. d. electrically charged. e. none of these.
Answer:
A molecule is a combination of two or more atoms. The correct option is B.
Explanation:
Molecules are the smallest particle of a chemical compound that are made up of two or more atoms which are held together by a chemical bond. The chemical bonds are usually formed due to sharing and transfer of electrons among the atoms. Examples of molecules in chemistry includes:
- water molecule ( H2O)
- table salt ( NaCl)
- CaCl
A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs.
Part A
What is the spring constant of each spring if the empty car bounces up and down 1.6 times each second?
Express your answer using two significant figures.in N/m.
Part B
What will be the car's oscillation frequency while carrying four 70 kg passengers?
Express in two sig figs in Hz.
Answer:
A) [tex]k=34867.3384\ N.m^{-1}[/tex]
B) [tex]\omega'\approx84\ Hz[/tex]
Explanation:
Given:
mass of car, [tex]m=1380\ kg[/tex]
A)
frequency of spring oscillation, [tex]f=1.6\ Hz[/tex]
We knkow the formula for spring oscillation frequency:
[tex]\omega=2\pi.f[/tex]
[tex]\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f[/tex]
[tex]\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6[/tex]
[tex]k_{eq}=139469.3537\ N.m^{-1}[/tex]
Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.
So, the stiffness of each spring is (as they are identical):
[tex]k=\frac{k_{eq}}{4}[/tex]
[tex]k=\frac{139469.3537}{4}[/tex]
[tex]k=34867.3384\ N.m^{-1}[/tex]
B)
given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:
[tex]\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }[/tex]
[tex]\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }[/tex]
[tex]\omega'\approx84\ Hz[/tex]
You are lost at night in a large, open field. Your GPS tells you that you are 122.0 m from your truck, in a direction 58.0o east of south. You walk 72.0 m due west along a ditch. How much farther, and in what direction, must you walk to reach your truck?
Answer:
The person is 187[m] farther and 70° south to east.
Explanation:
We can solve this problem by drawing a sketch of the location of the person and the truck, then we will draw the displacement vectors and finally the length of the vector and the direction of the vector will be measured in order to give the correct indication of where the person will have to move.
First we establish an origin of a coordinate system.
We can see in the attached schema that the red vector is the displacement vector from the last point to where the truck is located.
The length of the vector is 187 [m], and the direction is 70 degrees south to East.
This problem involves using vectors and trigonometry to calculate the direct distance and direction to the truck from the new location after walking due west. This is achieved by adding the horizontal and vertical components of the vectors representing the initial location and the walking path.
Explanation:To solve this, one may use vectors and trigonometry. Initially, you are 122.0m from your truck, 58.0 degrees east of south. This can be treated as a vector from your truck to your original location. Then, you walk 72.0m due west, which is another vector in the opposite direction.
To find the resulting vector, i.e., the direct distance and direction to the truck from your new location, we have to add these vectors. While the mathematics is somewhat complex, the concept involves adding the horizontal (east-west) and vertical (north-south) components of each vector. Once the resulting vector is calculated, the remaining distance to the truck can be found from its magnitude, and the direction from its angle relative to south.
This involves math calculations, including trigonometry and Pythagorean theorem.
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A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the gas (in J)
Answer:
Total work done in expansion will be [tex]3.60\times 10^5J[/tex]
Explanation:
We have given pressure P = 2.10 atm
We know that 1 atm [tex]=1.01\times 10^5Pa[/tex]
So 2.10 atm [tex]=2.10\times 1.01\times 10^5=2.121\times 10^5Pa[/tex]
Volume is increases from 3370 liter to 5.40 liter
So initial volume [tex]V_1=3.70liter[/tex]
And final volume [tex]V_2=5.40liter[/tex]
So change in volume [tex]dV=5.40-3.70=1.70liter[/tex]
For isobaric process work done is equal to [tex]W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J[/tex]
So total work done in expansion will be [tex]3.60\times 10^5J[/tex]
A rock is dropped from rest from a height h above the ground. It falls and hits the ground with a speed of 11 m/s. From what height should the rock be dropped so that its speed on hitting the ground is 22 m/s?
Answer:
Explanation:
The first part of question is about the height of the rock from which it falls and hit the ground with speed of 11 m/s. Lets find out that height.
We will use the formula,
[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]
As the initial velocity of the rock was zero. [tex]v_{f} = 0[/tex]
[tex]v^{2} _{f} = 2gh\\ h = v^{2} _{f} / 2g\\h = \frac{(11 m/s)^{2} }{2(9.8 m/s^{2} )} \\h = 6.17 m[/tex]
Now we have to find the height from which the rock should be dropped and it's speed on hitting the ground should be 22 m/s.
Again we will use the same formula, same calculation but the value of velocity now should be 22 m/s.
[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]
[tex]v^{2} _{f} = 2gh\\ h = \frac{(22m/s)^{2} }{2(9.8 m/s^{2}) } \\h = 24.69 m[/tex]
To double the speed of the rock when it hits the ground (from 11 m/s to 22 m/s), the height from which it is dropped should be quadrupled. Hence, the rock should be dropped from a height of 24.5 meters.
Explanation:The question is about finding the height from which a rock should be dropped so that its speed on hitting the ground is 22 m/s, given that when it is dropped from height h, its speed is 11 m/s. To solve this, we can use the physics equation for motion under constant acceleration, which is v² = 2gh, where v is the final velocity, g is the acceleration due to gravity, and h is the height of fall.
First, let us find the height h in the initial scenario: (11)² = 2*9.8*h => h = 6.125 m. Generally, the height h is proportional to the square of the speed, so if we double the final speed, the height should be quadrupled: h'(new height) = 4 * h = 4 * 6.125 m = 24.5 m.
Therefore, the rock should be dropped from a height of 24.5 m so that its speed on hitting the ground is 22 m/s.
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A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is acting down, how long does it take the proton to return to the horizontal plane?
Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2
[tex]T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}[/tex]
Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;
[tex]T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}[/tex]
T = 7.83 X10⁻⁷ s
Final answer:
The force exerted on a proton in a magnetic field can be calculated using the given formula with specified values.
Explanation:
To find the time it takes for the proton to return to the horizontal plane, we need to consider the motion in both the horizontal and vertical directions separately.
Given:
[tex]- Initial velocity (\(v_0\)) of the proton = \(3.0 \times 10^4\) m/s\\- Launch angle (\(\theta\)) = 30°\\- Electric field (\(E\)) = 400 N/C\\- We'll assume the gravitational acceleration (\(g\)) as \(9.8 \, \text{m/s}^2\) downward.[/tex]
Vertical Motion:
In the vertical direction, the proton undergoes uniformly accelerated motion under gravity.
Using the formula for vertical motion:
[tex]\[ v = u + at \] \\where: \\- \( v \) is the final velocity (which is 0 when the proton returns to the horizontal plane),\\- \( u \) is the initial vertical velocity,\\- \( a \) is the acceleration due to gravity,\\- \( t \) is the time taken.\\[/tex]
We can resolve the initial velocity into vertical and horizontal components:
[tex]\[ u_y = v_0 \sin(\theta) \]\[ u_y = 3.0 \times 10^4 \times \sin(30^\circ) \]\[ u_y \approx 3.0 \times 10^4 \times 0.5 \]\[ u_y = 1.5 \times 10^4 \, \text{m/s} \][/tex]
Now, let's find the time it takes for the vertical velocity to become zero:
[tex]\[ 0 = u_y - gt \]\[ t = \frac{u_y}{g} \]\[ t = \frac{1.5 \times 10^4}{9.8} \]\[ t \approx 1530.61 \, \text{s} \][/tex]
Horizontal Motion:
In the horizontal direction, there is no acceleration or deceleration acting on the proton. So, the time taken for horizontal motion is the same as the time taken for vertical motion.
The time it takes for the proton to return to the horizontal plane is approximately [tex]\( 1530.61 \) seconds.[/tex]
Think about the pencil-dropping activity that you did in the introduction. What did the target finally look like?
Answer:
By dropping a pencil from a certain fixed height again and again it will make the target super messay with marks of dot everywhere on the target and some even out side the target.
Explanation:
A yo-yo with a mass of 0.075 kg and a rolling radius of 2.50 cm (the distance from the axis of the pulley to where the string comes off the spool) rolls down a string with a linear acceleration of 6.50 m/s2. Approximate the rotational inertia of the yo-yo with that of disk with mass, m, and radius, r, rotating about its center (mr2/2). Calculate the tension in the string.
Answer:
0.24825 N
0.0000238701923077 kgm²
Explanation:
m = Mass of yo yo = 0.075 kg
a = Acceleration = 6.5 m/s²
g = Acceleration due to gravity = 9.81 m/s²
The net force is given by
[tex]F_n=mg-T[/tex]
[tex]\Rightarrow T=mg-ma[/tex]
[tex]\Rightarrow T=m(g-a)[/tex]
[tex]\Rightarrow T=0.075(9.81-6.5)[/tex]
[tex]\Rightarrow T=0.24825\ N[/tex]
The tension in the string is 0.24825 N
Angular acceleration is given by
[tex]\alpha=\dfrac{a}{r}\\\Rightarrow \alpha=\dfrac{6.5}{2.5\times 10^{-2}}\\\Rightarrow \alpha=260\ rad/s^2[/tex]
Torque is given by
[tex]\tau=I\alpha\\\Rightarrow Tr=I\alpha\\\Rightarrow I=\dfrac{Tr}{\alpha}\\\Rightarrow I=\dfrac{0.24825\times 2.5\times 10^{-2}}{260}\\\Rightarrow I=0.0000238701923077\ kgm^2[/tex]
The moment of inertia is 0.0000238701923077 kgm²
The tension in the string is equal to 0.2475 Newton.
Given the following data:
Mass of yo-yo = 0.075 kgRadius = 2.50 cm to m = [tex]\frac{2.5}{100} = 0.0025 \;m[/tex]Linear acceleration = 6.50 [tex]m/s^2[/tex]To determine the tension in the string:
First of all, we would determine the downward force applied by the yo-yo's weight:
[tex]F_y = mg[/tex]
Where:
[tex]F_y[/tex] is the yo-yo's weight. m is the mass of the yo-yo. g is acceleration due to gravity.
Substituting the given parameters into the formula, we have;
[tex]F_y = 0.075 \times 9.8\\\\F_y = 0.735 \; Newton[/tex]
Next, we would determine the force acting on the string:
[tex]F_s = 0.075 \times 6.5\\\\F_s = 0.4875\;Newton[/tex]
Now, we can find the tension in the spring:
[tex]Tension = F_y - F_s\\\\Tension = 0.735 - 0.4875[/tex]
Tension = 0.2475 Newton.
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