Answer:
a.-6.75 kgm/s
b.[tex]3375 N[/tex]
Explanation:
We are given that
Mass of baseball=0.15 kg
Initial speed=[tex]u=20m/s[/tex]
Final speed=[tex]v=-25m/s[/tex]
a.We know that
Impulse=Change in momentum=[tex]\Delta p=mv-mu=m(v-u)[/tex]
Momentum=[tex]mass\times velocity[/tex]
Using the formula
Impulse=[tex]0.15(-25-20)=-6.75 kgm/s[/tex]
b.Time=[tex]2\times 10^{-3} s[/tex]
Force=[tex]\frac{Impulse}{time}[/tex]
Using the formula
Average force exerted by the bat on the ball=[tex]\frac{-6.75}{2\times 10^{-3}}[/tex] N
Average force exerted by the bat on the ball=[tex]3375N[/tex]
Two uncharged metal balls, Y and X, each stand on a glass rod and are touching. A third ball carrying a negative charge, is brought near the first two While the positions of these balls are fixed, ball Y is connected to ground. Then the ground wire is disconnected. While Y and X remain in touch, the ball carring the negative charge is removed. Then ball Y and X are separated.
The phenomenon in question describes charging by induction. The presence of a negative charge induced a redistribution of charges in two uncharged metal balls, and grounding allowed excess charges to move away. The two uncharged balls ended up positively charged after the negatively charged ball was removed and the balls separated.
Explanation:The subject of this question relates to the concept of charging by induction, a common concept in Physics. When the third ball, which carries a negative charge, was brought near Ball Y and X, it caused the redistribution of charges in both balls due to the phenomenon of induction. As Ball Y was grounded, it provided a path for the electric charges to move away, thus becoming neutral again. However, Ball X retained the overall net positive charge due to the negative charge being near earlier.
After the ground wire is disconnected and the negatively charged ball removed, Ball Y and X still have net positive charges. This happens because the negative charge in the vicinity induced a redistribution of charges earlier, and when they were removed, there weren't any free charges nearby to neutralize this charge.
The separation of Ball Y and X made their charges apparent. Both balls are left with a positive charge as the excess electrons moved to the earth during grounding.
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Ball Y ends up with a positive charge, and Ball X with a negative charge due to induction caused by a nearby negatively charged ball. Grounding Ball Y neutralizes any negative charge, leaving it positively charged. The scenario exemplifies charging by induction.
Charging by Induction
This question deals with the concept of charging by induction in physics. Let's break down the scenario step-by-step:
Initially, Balls Y and X are both uncharged and in contact with each other.A third ball carrying a negative charge is brought close to Balls Y and X. Because of this, electrons on Ball Y will be repelled, moving to Ball X, resulting in Ball Y becoming positively charged and Ball X negatively charged.While the negatively charged ball is still near, Ball Y is connected to the ground. This allows the negative charge on Ball Y to be neutralized by the ground, leaving Ball Y positively charged.The ground connection is then removed. Now, Ball Y is positively charged and Ball X might still hold some negative charge due to the residual electrons.The negatively charged ball is then removed, followed by separating Balls Y and X.To address the specific parts of the question:
a. Charges on Each Ball
Ball Y will have a positive charge, and Ball X will have a negative charge. This distribution occurs due to the transfer of electrons from Y to X when the negatively charged third ball is near.
b. How the Balls Get These Charges
The negatively charged third ball induces a redistribution of charges in Balls Y and X. Electrons are repelled from Ball Y towards Ball X, leaving Ball Y positively charged and Ball X negatively charged.
c. Effect of Grounding Ball Y
When Ball Y is grounded while the negatively charged ball is close, the excess electrons on Ball Y will flow into the ground, neutralizing any negative charge and leaving it positively charged. When the ground connection is removed, Ball Y remains positively charged.
This process illustrates induction as the primary method of charging without direct contact.
u are holding the axle of a bicycle wheel with radius30 cm and mass 1.05 kg. You get the wheel spinning at arate of 77 rpm and then stop it by pressing the tire againstthe pavement. You notice that it takes 1.37 s for the wheelto come to a complete stop. What is the angular accelerationof the wheel
Answer:
-5.9 rad/s^{2}
Explanation:
radius (r) = 30 cm = 0.3 m
mass (m) = 1.05 kg
initial speed (u) = 77 rpm
final speed (v) = 0 rpm
time (t) = 1.37 s
angular acceleration =[tex]\frac{(final speed-initial speed)rad/s}{time}[/tex]
therefore
initial speed (U) = 77 rpm = 77 x (2π/60) = 8.06 rad/s
final speed (v) = 0 rpm = 0 rad/s
angular acceleration = [tex]\frac{0-8.06}{1.37}[/tex] = -5.9 rad/s^{2}
Answer:
-5.886 rad/s^2.
Explanation:
radius, r = 30 cm
= 0.3 m
mass, m
= 1.05 kg
initial speed, wo = 77 rpm
Converting from rpm to rad/s,
= 77 rpm * 2pi rad * 1 min/60 s
= 8.063 rad/s
final speed, wi = 0 rad/s
time, t = 1.37 s
angular acceleration = Δw/Δt
= (wi - wo)/t
= 8.063/1.37
= -5.886 rad/s^2.
Recall the observed behavior of the compass during the mapping of the magnetic field. Which of the following descriptions best matches the behavior you observed? Two correct answers required for full credit.
A. Where the field is strong the compass needle readily and quickly aligns with rapid oscillations around the equilibrium direction.
B. Where the field is strong the compass needle slowly oscillates around the final equilibrium position.
C. Where the field is weak the compass needle rapidly oscillates around the final equilibrium position.
D. Where the field is weak the compass needle slowly oscillates around the final equilibrium position.
E. Where the field is weak the compass needle readily and quickly aligns with rapid oscillations around the equilibrium direction.
Answer:
Option A & B
Explanation:
This is cause the compass needle is magnetized anointed in such a way that it responds to magnetic field strength.
You are driving to the grocery store at 18 m/s. You are 130 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.a. How far are you from the intersection when you begin to apply the brakes? b. What acceleration will bring you to rest right at the intersection? c. How long does it take you to stop?
Answer
given,
initial speed = 18 m/s
distance, d = 130 m
reaction time, t_r = 0.50 s
a) distance traveled in the reaction time
d= s x t_r
d=18 x 0.5
d = 9 m
distance of the traffic light, d = 130 - 9 = 121 m
b) deceleration
[tex]a = \dfrac{v^2-u^2}{2s}[/tex]
[tex]a = \dfrac{0^2-18^2}{2\times 121}[/tex]
a = -1.34 m/s²
c) Using equation of motion
v = u + a t
0 = 18 - 1.34 t
t = 13.43 s
Total time, T = 13.43 + 0.5 = 13.93 s
Part A. The distance from the intersection is 121 m.
Part B. The acceleration required is 1.34 m/s2
Part C. The total time to stop at the signal is 13.93 s.
How do you calculate the distance, acceleration and time?Given that the initial speed is 18 m/s. The distance is 130 m from an intersection when the traffic light turns red. The rejection time is 0.50 s.
Part A
The distance traveled in rejection time is given below.
distance traveled in rejection time = speed [tex]\times[/tex] rejection time
[tex]d_r = s\times t_r[/tex]
[tex]d_r = 18\times 0.50[/tex]
[tex]d_r = 9\;\rm m[/tex]
Hence the distance from the traffic light is given below.
[tex]d = 130 -d_r[/tex]
[tex]d = 130 -9[/tex]
[tex]d = 121\;\rm m[/tex]
The distance from the intersection is 121 m.
Part B
The acceleration can be calculated as given below.
[tex]a = \dfrac {v^2 - u^2}{2s}[/tex]
Where u is the initial speed, v is the final speed and s is the distance from the intersection.
[tex]a = \dfrac {0^2 - (18)^2}{2\times 121}[/tex]
[tex]a = - 1.34 \;\rm m/s^2[/tex]
The acceleration required is 1.34 m/s2. Here negative sign shows that this is deacceleration.
Part C
The time to stop at the signal can be calculated as given below.
[tex]v = u +at[/tex]
[tex]0 = 18 - 1.34 \times t[/tex]
[tex]t = \dfrac {18}{1.34}[/tex]
[tex]t = 13.43 \;\rm s[/tex]
Total Time = [tex]13.43 + 0.5[/tex]
Total Time = 13.93 s
Hence the total time to stop at the signal is 13.93 s.
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answers If visible light and radio waves are both examples of electromagnetic waves, why can’t we see radio waves?
Answer:
Radio Waves:
Radio waves being the lowest-energy form of light and are known to be produced by electrons spiraling around magnetic fields. These Magnetic fields are generated by stars, including our sun, and many weird celestial objects like black holes and neutron stars.
Explanation:
All electromagnetic radiation is light, but we can only see a small portion of this radiation that is the portion we call visible light. Cone-shaped cells in our eyes act as receivers tuned to the wavelengths in this narrow band of the spectrum
Visible light and radio waves are both examples of electromagnetic waves, but visible light has shorter wavelengths that our eyes can detect, while radio waves have longer wavelengths that are outside the range of what our eyes can see.
Explanation:Visible light and radio waves are both examples of electromagnetic waves, but they have different wavelengths. Visible light has a shorter wavelength than radio waves, which is why we can see it. Our eyes are sensitive to the wavelengths of visible light, but not to the longer wavelengths of radio waves.
The human eye can detect wavelengths within a certain range, known as the visible spectrum, which includes the colors of the rainbow. Radio waves have much longer wavelengths, ranging from meters to kilometers. These longer wavelengths are outside the range of what our eyes can detect, so we cannot see radio waves.
However, even though we can't see radio waves, we can still use them for various purposes, such as wireless communication, radar, and satellite transmission.
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A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble during the same amount of time. Which object has the greatest value for each of these quantities?
Answer: pebble has the greatest value of acceleration due to its low mass.
Explanation:
According to newton's second law of motion,
Force = mass × acceleration
For boulder,
Acceleration = Force/mass
Force acting on boulder is 200N
Its mass is 100kg
Acceleration = 200/100
Acceleration = 2m/s²
Similarly for pebble,
Force on pebble = 200N (the same as boulder)
Its mass is 130g = 0.13kg(has to be converted to the standard unit which is kg)
Its acceleration = 200/0.13
Acceleration of pebble = 1538.5m/s²
Since the question doesn't specify what to compare, we will compare their accelerations.
Therefore, pebble has the greatest value of acceleration due to its low mass.
Xenon fluoride can be prepared by heating a mixture of Xe and F2 gases to a high temperature in a pressure-proof container. Assume that xenon gas was added to a 0.25 liter container until its pressure reached 0.12 atm at 0.0°C. Fluorine gas was then added until the total pressure reached 0.72 atm at 0.0°C. After the reaction was complete, the xenon was consumed completely, and the pressure of the F2 remaining in the container was 0.36 atm at 0.0°C. What is the empirical formula of the xenon fluoride?
Answer:
XeF₄
Explanation:
Dalton law's of partial pressure states that the total pressure equal to the sum of the partial pressure of the individual gases that make up the mixture.
Pt = Pxe + Pf₂
0.72 = 0.12 + Pf₂
0.72 - 0.12 = Pf₂
0.60 atm = Pf₂
after the the reaction was complete, the pressure of F₂ remaining = 0.36 atm
pressure of the consumed F₂ = 0.60 - 0.36 = 0.24 atm
Pxe / total pressure = number of mole / total number of mole of gas present
0.24 / 0.72 = nf / nt
0.12 / 0.72 = nxe / nt
comparing the two
(1/ 3) / ( 1/6) = (nf/ nt) / ( nxe/ nt)
nf / nxe = 2 / 1
the emperical formula = XeF₄
The empirical formula of the xenon fluoride is :
-XeF₄
"Dalton law's of partial pressure"It states that the Total pressure break even with to the whole of the fractional weight of the person gasses that make up the blend.
Pt = Pxe + Pf₂0.72 = 0.12 + Pf₂0.72 - 0.12 = Pf₂0.60 atm = Pf₂Pressure of F₂ remaining = 0.36 atm
Pressure of the consumed F₂ = 0.60 - 0.36 = 0.24 atm
Pxe / total pressure = number of mole / total number of mole of gas present
0.24 / 0.72 = nf / nt
0.12 / 0.72 = nxe / nt
comparing the two:
(1/ 3) / ( 1/6) = (nf/ nt) / ( nxe/ nt)
nf / nxe = 2 / 1
The emperical formula = XeF₄
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A hammer taps on the end of a 4.0-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pulses of sound, one that travels through the metal and one that travels through the air. The pulses are separated in time by 11.0 ms.
1. What is the speed of sound in this metal? v = 6060 m/s
2.If the bar was twice as long, what would be the time difference?
Answer: c*m on now
Explanation:
u dumb b***h
Suppose that the voltage of the battery in the circuit is 3.3 V, the magnitude of the magnetic field (directed perpendicularly into the plane of the screen) is 0.66 T, and the length of the rod between the rails is 0.19 m. Assuming that the rails are very long and have negligible resistance, find the maximum speed attained by the rod after the switch is closed.
Answer:
Maximum velocity of rod will be equal to 26.315 m/sec
Explanation:
We have given voltage of the battery in the circuit V = 3.3 volt
Magnetic field B = 0.66 Tesla
Length of the rod l = 0.19 m
We know that emf is given by e = BVl
We have to find the maximum velocity of the rod
Here velocity is maximum
So [tex]e=v_{max}Bl[/tex]
So [tex]3.3=v_{max}\times 0.66\times 0.19[/tex]
[tex]v_{max}=26.315m/sec[/tex]
So maximum velocity of rod will be equal to 26.315 m/sec
A 170 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm against the spring, then released. A student with a stopwatch finds that 14 oscillations take 11.0 s. What is the spring constant?
Answer:k = 10.83 N/m²
Explanation: The angular frequency (ω), spring constant (k) and mass is related by the formulae below
ω = √k/m
But ω = 2πf, where f = frequency.
f = number of oscillations /time taken
Number of oscillations = 14, time taken = 11s
f = 14/11 = 1.27Hz.
ω = 2×22/7×1.27
ω = 7.98 rad/s.
By substituting this parameters into ω = √k/m
Where ω = 7.98rad/s, m = 170g = 170/1000 = 0.17kg.
7.98 = √k/0.17
By squaring both sides
(7.98)² = k/ 0.17
k = (7.98)² × 0.17
k = 10.83 N/m²
Final answer:
To find the spring constant (k) of the spring system, we use the period of oscillation formula. After calculating the period of one oscillation (T = 0.7857 s), we solve for k = m/(T/2π)^2 and find that k is approximately 44.1 N/m.
Explanation:
The student asked how to find the spring constant of a spring if a 170 g air-track glider is attached to it, pushed in 11.2 cm, released, and makes 14 oscillations in 11.0 seconds. The spring constant (k) can be found using the formula for the period of a mass-spring system (T = 2π√(m/k)) where m is mass and k is the spring constant.
First, calculate the period of one oscillation by dividing the total time by the number of oscillations: T = 11.0 s / 14 = 0.7857 s. Then rearrange the period formula to solve for k: k = m/(T/2π)2. Convert the mass to kilograms (m = 0.170 kg) and substitute the values to calculate k.
Doing the calculation: k = 0.170 kg / (0.7857 s / 2π)2 gives us the spring constant k. Upon solving, we find that the spring constant is approximately 44.1 N/m.
A flea jumps straight up to a maximum height of 0.490 m . How long is the flea in the air from the time it jumps to the time it hits the ground?
Answer:
0.62 s
Explanation:
given,
maximum height of the flea = 0.49 m
velocity at maximum height = 0 m/s
now, calculating initial velocity
using equation of motion
[tex]v^2 = u^2 + 2 g h[/tex]
[tex]0^2 = u^2 - 2\times 9.8 \times 0.49[/tex]
[tex]u^2 = 9.604[/tex]
[tex]u = 3\ m/s[/tex]
now, calculating time he take to reach at the highest point
v = u + g t
0 = 3 - 9.8 x t
9.8 t = 3
t = 0.31 s
Time it will be in air will be twice the time it took to reach to the maximum height.
Time for which it was in air = 2 x 0.31 s = 0.62 s
The energy released by the exploding gunpowder in a cannon propels the cannonball forward. Simultaneously, the cannon recoils. The mass of the cannonball is less than that of the cannon. Which has the greater kinetic energy, the launched cannonball or the recoiled cannon? Assume that momentum conservation applies.
Answer:
the launched cannonball
Explanation:
In a cannon firing a cannonball, while the total momentum is conserved, the cannonball has greater kinetic energy than the recoiling cannon due to its higher velocity and the relationship between kinetic energy and velocity.
The subject of this question is Physics, specifically dealing with momentum conservation and kinetic energy in the context of a cannon firing a cannonball. The law of conservation of momentum states that the total momentum before and after an event is constant if no external forces are acting on the system. When a cannon fires a cannonball, the momentum is conserved, meaning the cannon and the cannonball together possess the same total momentum post-explosion as the stationary cannon had before firing (which is zero). Though both the cannon and the cannonball have momentum of equal magnitude but opposite direction, the distribution of kinetic energy between them is not equal.
Kinetic energy, which depends on both mass and the square of velocity, will not be conserved in the same way as momentum. The kinetic energy is given by the relation KE = (1/2)mv2. Considering the masses of the two objects, a lighter cannonball and a heavier cannon, if both the cannon and cannonball have the same magnitude of momentum, the cannonball will have a much higher velocity due to its smaller mass. Since kinetic energy depends on the square of velocity, the cannonball, with a higher velocity, will have greater kinetic energy than the heavier, slower-moving cannon. Thus, even though momentum is conserved, the kinetic energies will differ, with the cannonball gaining more kinetic energy than the recoiling cannon.
In summary, the law of conservation of energy applies, but kinetic energy is not conserved like momentum. The chemical energy stored in the gunpowder is converted into heat and the kinetic energy of both the cannon and the cannonball. Although the cannon has a larger mass, it will recoil with less velocity than the fired cannonball, which means the cannonball will have a greater kinetic energy.
How high would you need to lift a 2 kg bottle of soda to increase its potential energy by 50 Joules? (use G=10 N/KG.)
Answer:
2.5m
Explanation:
PE = MGH
50 = 2*10*H
50 = 20H
H = 2.5m
A small pumpkin is launched from a very versatile catapult at a variety of speeds and angles. The angles are measured relative to the horizontal. Rank each of the following trials by the maximum height reached by the pumpkin. (1=highest) Consider air resistance to be negligible.
1 2 3 4 Speed: 30.7m/s, Angle: 35.7°
1 2 3 4 Speed: 47.9m/s, Angle: 33.2°
1 2 3 4 Speed: 21.8m/s, Angle: 67.7°
1 2 3 4 Speed: 7.7m/s, Angle: 62.1°
Answer:
Explanation:
Maximum height of a projectile
[tex]H_{max}=\frac{u^2\sin ^2\theta }{2g}[/tex]
(1)[tex]u=30.7\ m/s[/tex]
[tex]\theta =35.7^{\circ}[/tex]
[tex]H_1=\frac{30.7^2\times \sin ^2(35.7)}{2\times 9.8}[/tex]
[tex]H_1=16.37\ m[/tex]
(2)[tex]u=47.9\ m/s[/tex]
[tex]\theta =33.2^{\circ}[/tex]
[tex]H_2=\frac{47.9^2\times \sin ^2(33.2)}{2\times 9.8}[/tex]
[tex]H_2=35.09\ m[/tex]
(3)[tex]u=21.8\ m/s[/tex]
[tex]\theta =67.7^{\circ}[/tex]
[tex]H_3=\frac{21.8^2\times \sin ^2(67.7)}{2\times 9.8}[/tex]
[tex]H_3=20.75\ m[/tex]
(4)[tex]u=7.7\ m/s[/tex]
[tex]\theta =62.1^{\circ}[/tex]
[tex]H_4=\frac{7.7^2\times \sin ^2(62.1)}{2\times 9.8}[/tex]
[tex]H_4=2.36\ m[/tex]
Therefore height reached by pumpkin can be arranged as
[tex]H_4<H_1<H_3<H_2[/tex]
Final answer:
The maximum height reached by the pumpkin depends on the vertical component of the initial velocity, which is highest for the trial with a speed of 21.8m/s and angle of 67.7 degrees and lowest for the speed of 7.7m/s at an angle of 62.1 degrees.
Explanation:
To rank the maximum height reached by the pumpkin in each trial, we can rely on the principle of projectile motion. The maximum height depends primarily on the vertical component of the initial velocity and can be compared by isolating this component using the initial speeds and angles given.
Using the formula for the vertical component of velocity: [tex]Vy = V * sin(\(\theta\)),[/tex] where V is the initial speed and [tex]\(\theta\)[/tex] is the launch angle. Then, the maximum height H can be roughly compared using [tex]H \(\propto Vy^2[/tex], assuming gravitational acceleration is constant and air resistance is negligible.
Speed: 21.8m/s, Angle: [tex]67.7\(\deg\)[/tex] - Highest vertical componentSpeed: 47.9m/s, Angle: [tex]33.2\(\deg\)[/tex]Speed: 30.7m/s, Angle: [tex]35.7\(\deg\)[/tex]Speed: 7.7m/s, Angle: [tex]62.1\(\deg\)[/tex] - Lowest vertical componentIf you were to drill a hole into the wooden sphere, and fill it with lead, how much of the volume of the wood would have to be replaced to make the whole sphere neutrally buoyant (i.e. have the same density as water)?
Answer:
[tex]\displaystyle V_l=V\ \frac{\rho_o -\rho_w }{\rho_l-\rho_w}[/tex]
Explanation:
Density
It's a physical magnitude that relates the mass of an object with the volume it occupies. The formula is
[tex]\displaystyle \rho = \frac{m}{V}[/tex]
Equivalently
[tex]m=\rho V[/tex]
The density of water is 1 gr/ml, 1 Kg/lt, 1000 kg/m^3 or any equivalent unit.
Assume the wood sphere has a volume V and a density [tex]\rho_w[/tex], thus
[tex]m_w=\rho_w V[/tex]
A hole is to be drilled inside the wood so it means part of its volume will be filled with lead, which mass will be
[tex]m_l=\rho_l V_l[/tex]
The volume of wood is the total volume V minus the (unknown) volume of lead, thus
[tex]V_w=V-V_l[/tex]
The total mass of the modified sphere is
[tex]m_s=m_w+m_l=\rho_w V_w+\rho_l V_l[/tex]
Substituting Vw
[tex]m_s=m_w+m_l=\rho_w (V-V_l)+\rho_l V_l[/tex]
Operating
[tex]m_s=m_w+m_l=\rho_w V-\rho_w V_l+\rho_l V_l[/tex]
[tex]m_s=m_w+m_l=\rho_w V+ V_l(\rho_l -\rho_w)[/tex]
The total volume of the sphere doesn't change, which means
[tex]V_s=V[/tex]
The new density of the modified sphere is
[tex]\displaystyle \rho_s = \frac{m_s}{V}[/tex]
[tex]\displaystyle \rho_s = \frac{\rho_w V+ V_l(\rho_l -\rho_w)}{V}[/tex]
This density must be equal to the density of water [tex]\rho_o[/tex]
[tex]\displaystyle \rho_o = \frac{\rho_w V+ V_l(\rho_l -\rho_w)}{V}[/tex]
Operating and solving for vl
[tex]\displaystyle V_l=\frac{\rho_o V-\rho_w V}{\rho_l-\rho_w}[/tex]
Or equivalently
[tex]\displaystyle \boxed{V_l=V\ \frac{\rho_o -\rho_w }{\rho_l-\rho_w}}[/tex]
The equation would be enough for the volume of wood to keep the sphere neutrally buoyant would be [tex]Vt= Vp0-Pw/Pt-Pw[/tex]
Calculations and Parameters:The formula for the physical magnitude is p= m/v
Given that the density of water is 1 gr/ml, 1 Kg/lt, 1000 kg/m^3 or any equivalent unit.
We would assume the wood sphere has a volume V and a density Pw,
[tex]Mw= PwV[/tex]
Since a hole is to be drilled inside the wood so it means part of its volume will be filled with lead, the mass will be [tex]Mt= PtVt[/tex]
The volume of wood is the total volume V minus the (unknown) volume of lead, thus
[tex]Vw= V- Vt[/tex]
Hence, after further evaluation, the density would be [tex]Vt= V Po- Pw/Pt- Pw[/tex]
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What is the relationship between the amount of stretch or compression in an elastic object and the amount of force that is applied to that object?
Applied force is one-half the amount of stretch or compression.
They are directly proportional.
They are inversely proportional.
Applied force is double the amount of stretch or compression.
The relationship between the amount of stretch or compression in an elastic object and the amount of force that is applied to that object is directly proportional to each other.
Answer:
Option B
Explanation:
Ideal elastic objects are those which will retain its original form after the removal of external force on it. So this specialty of retaining their original form after deforming due to external force is termed as elasticity. And the objects exhibiting this property is termed as elastic object. The most important law in the field of elasticity which is obeyed by all elastic object is the Hooke's law. As per Hooke's law, the external force applied on any object for deformation will be directly proportional to the changes or deformation exhibited by the object due to force. In other or simple words, the strain produced due to the external force is directly proportional to the stress acting on that object.
Stress ∝ Strain
Stress is the measure of force applied on that object and strain is the measure of amount of stretch or compression due to stress or applied force.
So option B is the correct option.
Thus, the relationship between the amount of stretch or compression in an elastic object and the amount of force that is applied to that object is directly proportional to each other.
The amount of stretch or compression in an elastic object is directly proportional to the force applied, based on the principle known as Hooke's Law. This relationship is linear, meaning that changes in force result in proportional changes in deformation.
Explanation:The amount of stretch or compression in an elastic object is directly proportional to the amount of force that is applied. This principle is captured by Hooke's Law, which states when you apply force to stretch or compress a spring, the resulting spring force is proportional, and in the opposite direction, to the displacement or deformation from its position of equilibrium. Essentially, if you double the force, the amount of stretch or compression doubles, and if you halve the force, the stretch or compression is halved as well. This relationship can be represented by the equation F = -kx, where F is the force, k is the spring's force constant, and x is the displacement from the equilibrium position.
Tension and compression play a role here, where tension forces stretch the object while compression forces push the object together. Hooke's Law applies to ideal springs and other elastic objects provided the deformation does not exceed the material's elastic limit.
Which of Galileo's observations directly disproved Ptolemy's epicycle model of the Solar System, showing that the Sun is at the center and not Earth?
Answer:
Venus observation.
Explanation:
Galileo had learned regarding the heliocentric (Sun-centered) idea of Copernicus, and acknowledged it. However, the theory was proven by Galileo's observations of Venus. Galileo concluded that Venus should travel round the Sun, sometimes passing behind and then beyond, instead of directly rotating around the Earth.
1. Our entire solar system orbits around the center of the Milky Way Galaxyabout once every 230 million years.
2. The Milky Way and Andromeda galaxies are among a few dozen galaxies that make up our Local Group.
3. The Sun appears to rise and set in our sky because Earth orbitsonce each day.
4. You are one year older each time Earth rotatesabout the Sun.
5. On average, galaxies are getting farther apart with time, which is why we say our universeis expanding.
6. Our solar systemis moving toward the star Vega at about 70,000 km/hr.
Answer: Hello! Apparently, your question is incomplete. Those were sentences in which you had to complete with missing information, so here we go:
1- Our entire solar system orbits around the center of the MILKY WAY GALAXY about once every 230 million years.
2- The Milky Way and Andromeda galaxies are among a few dozen galaxies that make up our LOCAL GROUP.
3 - The Sun appears to rise and set in our sky because Earth ROTATES once each day.
4 - You are one year older each time Earth ORBITS about the Sun.
5 - On average, galaxies are getting farther apart with time, which is why we say our UNIVERSE is expanding.
6 - Our SOLAR SYSTEM is moving toward the star Vega about 70,000 km/hr.
Careers in this career cluster also include related professional and technical support activities such as production planning and control, maintenance and manufacturing/process engineering.A. True
B. False
The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound level goes up by 26.0 dB?
Answer:
The amplitude of wave increases by a factor of 20
Explanation:
The Sound level in decibel β is described as
[tex]\beta =(10dB)log(\frac{I}{I_{o} })[/tex]
Where I₀=10⁻¹²W/m² is reference intensity
Suppose β₁ is initial sound intensity and β₂ is final sound intensity.Therefore the change in sound intensity is given as
β₁-β₂=26.0 dB
[tex](10dB)log(\frac{I_{2} }{I_{o}} )-(10dB)log(\frac{I_{1} }{I_{o}} )=26.0dB\\(10dB)[log(\frac{I_{2} }{I_{o}} )-log(\frac{I_{1} }{I_{o}} )]=26.0dB\\log(\frac{\frac{I_{2}}{I_{o}} }{\frac{I_{1}}{I_{o}} })=\frac{26dB}{10dB}\\ log\frac{I_{2}}{I_{1}}=2.6\\ \frac{I_{2}}{I_{1}}=10^{2.6}[/tex]
The intensity of sound is directly proportional to the amplitude of wave squared
So
I∝A²
[tex]\frac{I_{1}}{I_{2}}=(\frac{A_{1}}{A_{2}})^{2}\\ (\frac{A_{1}}{A_{2}})=\sqrt{(\frac{I_{1}}{I_{2}})}\\ (\frac{A_{2}}{A_{1}})=\sqrt{(\frac{I_{2}}{I_{1}})}\\ (\frac{A_{2}}{A_{1}})=\sqrt{10^{2.6} }\\ (\frac{A_{2}}{A_{1}})=20[/tex]
Therefore the amplitude of wave increases by a factor of 20
According to the animation, in the first half of the lunar cycle, the waxing, bright part of the moon appears on the ____ side of the Earth. After full moon when the moon is waning, the illuminated part of the moon appears on the ______ side.
Answer: Blanks in order are: Right, Left
Explanation:
The moon rotates around the Earth. It completes its rotation in roughly 29.5 days which is known as the lunar cycle. The first half of rotation is known as waxing and in this the moon starts to become a full moon so bright side of the moon appears on the left side of Earth. After this ends, the waning part starts which means the decreasing of light of moon so bright side is on the left side of the Earth.
During the first half of the lunar cycle, the bright part of the moon (waxing moon) appears on the right side and after the full moon, the lit part of the moon (waning moon) appears on the left side.
Explanation:According to the animation, in the first half of the lunar cycle, the waxing, or growing, bright part of the moon appears on the right side (from an observer in the northern hemisphere). This phase is known as a waxing moon. After the full moon when the moon starts to shrink, or wane, the illuminated part of the moon appears on the left side. This phase is known as a waning moon.
Learn more about Lunar CycleFrom the following list of words, choose the correct one to complete the statement: absorption, reflection, emission, transmission.
a. Light coming from your computer screen is an example of _________________.
b. Yellow light hitting a yellow banana is an example of _______________.
c. Blue light hitting a yellow banana is an example of _______________.
d. Sunlight passing through a glass door is an example of ______________.
Answer:
a) Emission
b) Reflection
c) Absorption
d) Transmission
Explanation:
a) The screen of computers (most of them LED or LCD) are a bunch of tiny lights that take power of our pc to produce light so they EMMIT light.
b) Sun's light contains all the visible colors of electromagnetic spectrum, when sun light hits an object, due their chemical characteristics the object absorbs a bunch of wavelengths and reflects others, the reflected light is the color we perceive with our eyes. So, for a banana, it absorbs almost all the other colors and reflects only yellow light.
c) As explained on b) a banana absorbs all the other colors except yellow so blue is absorbed on it.
d) When a light hits transparent medium, some rays are reflected and other passes through the medium so that rays are called transmitted and that is the case of Sunlight passing through a glass door.
Final answer:
Correct words are emission for light from a computer screen, reflection for yellow light on a banana, absorption for blue light on a banana, and transmission for sunlight through a glass door. These terms describe how light interacts with different materials through emission, reflection, absorption, and transmission.
Explanation:
From the list of terms provided, the correct words to complete the statements about the behavior of light are:
a. Light coming from your computer screen is an example of emission.
b. Yellow light hitting a yellow banana is an example of reflection.
c. Blue light hitting a yellow banana is an example of absorption.
d. Sunlight passing through a glass door is an example of transmission.
When we talk about light waves interacting with materials, we discuss concepts such as reflection, absorption, and transmission. Reflection occurs when light bounces off an object, such as a yellow light reflecting off a yellow banana. Absorption occurs when an object takes in the light energy, which might later be transformed into other forms of energy like heat, as is the case when a yellow banana absorbs blue light. Lastly, transmission refers to the passage of light through a transparent or semi-transparent material, such as sunlight passing through a glass door.
A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= α t2− β t3, where α = 1.53 m/s2 and β = 4.80×10−2 m/s3. Calculate the average velocity of the car for each time interval:_______.
a) t = 0 to t = 2.00 s.
b) t = 0 to t = 4.00 s.
c) t = 2.00 s to t = 4.00 s.
Answer:
(a) [tex]V_{avg}=2.868m/s[/tex]
(b) [tex]V_{avg}=5.352m/s[/tex]
(c) [tex]V_{avg}=7.836m/s[/tex]
Explanation:
Given data
x(t)=αt²-βt³
α=1.53m/s²
β=0.0480m/s³
First we need to find distance x at these time so
x(t)=1.53t²-0.0480t³
at t=0
x(0)=1.53(0)²-0.0480(0)³=0m
at t=2
x(2)=1.53(2)²-0.0480(2)³=5.736m
at t=4s
x(4)=1.53(4)²-0.0480(4)³=21.408 m
For(a) Average velocity at t=0s to t=2s
The average velocity is given as
Vavg=Δx/Δt
[tex]V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(0)=0m\\x(2)=5.736m\\V_{avg}=\frac{5.736m-0m }{2s-0s}\\V_{avg}=2.868m/s[/tex]
For(b) Average velocity at t=0s to t=4s
The average velocity is given as
Vavg=Δx/Δt
[tex]V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(0)=0m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-0m }{4s-0s}\\V_{avg}=5.352m/s[/tex]
For(c) Average velocity at t=2s to t=4s
The average velocity is given as
Vavg=Δx/Δt
[tex]V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(2)=5.736m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-5.736m }{4s-2s}\\V_{avg}=7.836m/s[/tex]
Final answer:
The average velocity of the Honda Civic over the time intervals from t=0 to t=2.00 s is 2.868 m/s, from t=0 to t=4.00 s is 5.352 m/s, and from t=2.00 s to t=4.00 s is 7.836 m/s, calculated using the given equation and the variables α and β.
Explanation:
To find the average velocity of the Honda Civic over specific time intervals using the equation x(t) = α t^2 - β t^3, where α = 1.53 m/s^2 and β = 4.80×10^-2 m/s^3, we must calculate the change in position (Δx) over the change in time (Δt) for each interval.
a) t = 0 to t = 2.00 s
Initial position, x(0) = 0 (since any term with t = 0 results in 0).
Final position, x(2) = 1.53×2^2 - 4.80×10^-2×2^3 = 6.12 - 0.384 = 5.736 m.
Δx = 5.736 - 0 = 5.736 m, Δt = 2 s.
Average velocity = Δx/Δt = 5.736 m / 2 s = 2.868 m/s.
b) t = 0 to t = 4.00 s
Final position, x(4) = 1.53×4^2 - 4.80×10^-2×4^3 = 24.48 - 3.072 = 21.408 m.
Δx = 21.408 - 0 = 21.408 m, Δt = 4 s.
Average velocity = Δx/Δt = 21.408 m / 4 s = 5.352 m/s.
c) t = 2.00 s to t = 4.00 s
Using the previously calculated final positions for t = 2 and t = 4:
Δx = 21.408 - 5.736 = 15.672 m, Δt = 4 s - 2 s = 2 s.
Average velocity = Δx/Δt = 15.672 m / 2 s = 7.836 m/s.
Austin left the park traveling 4 mph. Then, 3 hours later, Wyatt left traveling the same direction at 10 mph. How long until Wyatt catches up with Austin?
ANSWER: 2 hours
EXPLANATION: After three hours Austin has traveled 12 miles. Wyatt will start up and in an hour will be at 10 miles, but by that time Austin will be at 16 miles. One more mile and Wyatt will be at 20 miles and so will Austin. So that would make the answer 2 hours.
Which of the following best summarizes the relationship between dehydration reactions and hydrolysis?
-Dehydration reactions can occur only after hydrolysis.
-Dehydration reactions ionize water molecules and add hydroxyl groups to polymers; hydrolysis reactions release hydroxyl groups from polymers.
- Dehydration reactions assemble polymers, and hydrolysis reactions break down polymers.
-Hydrolysis creates monomers, and dehydration reactions break down polymers.
-Dehydration reactions eliminate water from lipid membranes, and hydrolysis makes lipid membranes water permeable.
Answer:
Dehydration reactions assemble polymers, and hydrolysis reactions break down polymers.
Explanation:
dehydration reaction is a conversion that involves the loss of water from the reacting molecule or ion.
Hydrolysis is defined as any chemical reaction in which a molecule of water ruptures one or more chemical bonds.
Dehydration reactions link monomers together into polymers by releasing water, and hydrolysis breaks polymers into monomers using a water molecule. Monomers are just single unit molecules and polymers are chains of monomers.
Dehydration reactions and hydrolysis have opposite roles in relation to polymers. Dehydration reactions involve the removal of a water molecule to assemble polymers, while hydrolysis involves the addition of a water molecule to break down polymers.
Explanation:The relationship between dehydration reactions and hydrolysis is that they essentially have opposite roles in relation to polymers. Dehydration reactions are processes that involve the removal of a water molecule, which allows for the formation or assembly of polymers. This occurs because a hydroxyl group (OH) is removed from one molecule and a hydrogen (H) from another, enabling the two molecules to bond and form a polymer.
In contrast, hydrolysis is a reaction that involves the addition of a water molecule, which results in the breakdown of polymers into monomers. During hydrolysis, a water molecule is split into a hydroxyl group and a hydrogen, and these are used to break the bonds within a polymer, resulting in monomers.
So, to summarize, the best choice that describes the relationship between these two processes is: Dehydration reactions assemble polymers, and hydrolysis reactions break down polymers.
Learn more about Dehydration reactions and hydrolysis here:https://brainly.com/question/34331091
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You are standing in an elevator that is accelerating upward at 2 m/s2. What is the normal force acting on you by the elevator if your mass is 70 kg?
Answer:
784 N.
Explanation:
The expression for the Normal reaction acting on a body inside an elevator that is moving upward is given as
R = m(g+a)................. Equation 1
Where R = normal reaction, g = acceleration due to gravity, a = acceleration, m = mass .
Given: m = 70 kg, a = 2 m/s²
Constant: g = 9.8 m/s²
Substitute into equation 1
R = 70(2+9.2)
R = 70(11.2)
R = 784 N.
Hence the normal force = 784 N.
If two identical conducting spheres are in contact, any excess charge will be evenly distributed between the two. Three identical metal spheres are labeled AA, BB, and CC. Initially, AA has charge qq, BB has charge −q/2−q/2, and CC is uncharged.
Answer:
The solution is in the attached files below
Explanation:
A train travels due south at 25 m/s (relative to the ground) in a rain that is blown toward the south by the wind. The path of each raindrop makes an angle of 66° with the vertical, as measured by an observer stationary on the ground. An observer on the train, however, sees the drops fall perfectly vertically. Determine the speed of the raindrops relative to the ground.
Answer:
Explanation:
Given
Train travels towards south with a velocity if [tex]v_t=25\ m/s[/tex]
Rain makes an angle of [tex]\theta =66^{o}[/tex] with vertical
If an observer sees the drop fall perfectly vertical i.e. horizontal component of rain velocity is equal to train velocity
suppose [tex]v_r[/tex] is the velocity of rain with respect to ground then
[tex]v_r\sin\theta =v_t[/tex]
[tex]v_r\times \sin (66)=25[/tex]
[tex]v_r=27.36\ m/s[/tex]
Therefore velocity of rain drops is 27.36 m/s
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1050 kg and was approaching at 6.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.
Answer:
v = 11.0 m/s at 198.6° (18.6° south of west)
ΔKE = -145 kJ
Explanation:
I assume you want to find the final velocity and the change in kinetic energy.
Take east to be +x and north to be +y.
Momentum is conserved in the x direction:
(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ
vₓ = -10.4 m/s
Momentum is conserved in the y direction:
(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ
vᵧ = -3.50 m/s
The magnitude of the final velocity is:
v² = (-10.4 m/s)² + (-3.50 m/s)²
v = 11.0 m/s
The direction of the final velocity is:
θ = atan(-3.50 m/s / -10.4 m/s)
θ = 198.6°
The initial kinetic energy is:
KE₀ = ½ (1050 kg) (6.00 m/s)² + ½ (750 kg) (25.0 m/s)²
KE₀ = 253,275 J
The final kinetic energy is:
KE = ½ (1800 kg) (11.0 m/s)²
KE = 108,682 J
The change in kinetic energy is:
ΔKE = 108,682 J − 253,275 J
ΔKE ≈ -145,000 J
"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its speed when it passes the equilibrium position."
Answer:
1.3m/s
Explanation:
Data given,
Mass,m=1.0kg,
Amplitude,A=0.10m,
Frequency,f=2.0Hz.
From the equation of a simple harmonic motion, the displacement of the object at a given time is define as
[tex]x=Acos\alpha \\[/tex]
we can express the velocity by the derivative of the displacement,
Hence
[tex]V=-Awsin\alpha \\[/tex]
at equilibrium, the velocity becomes
[tex]V=wA\\w=2\pi f[/tex]
Hence if we substitute values we arrive at
[tex]V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s[/tex]
Answer:
haha lol
Explanation:
sorry but i dont know this