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3 years ago

If the amplitude of a simple harmonic oscillator is doubled, by what factor does the total energy increase?

Physics

1 answer:

dybincka [34]3 years ago

6 0

Answer:

c)by a factor of four

Explanation:

The total energy of a simple harmonic oscillator is given by

$E=\frac{1}{2}kA^2$

where

k is the spring constant of the oscillator

A is the amplitude of the motion

In this problem, the amplitude of the oscillator is doubled, so

A' = 2A

Therefore, the new total energy is

$E'=\frac{1}{2}k(2A)^2=4(\frac{1}{2}kA^2)=4E$

So, the total energy increases by a factor 4.

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A car traveling 20 m/s to the north collides with a car traveling 10 m/s to the

sergiy2304 [10]

Answer: C

Explanation:

In collision, whether elastic or inelastic collisions, momentum is always conserved. That is, the momentum before collision will be equal to the momentum after collision.

Change in momentum of the system will be momentum after collision minus total momentum before collision.

Since momentum is a vector quantity, the direction will also be considered.

Momentum = MV - mU

Let

M = 800 kg is going north

at V = 20 m/s and the other car

m= 800 kg is going south

at U = 10m/s.

Substitute all the parameters into the formula

Momentum = (800 × 20) - (800 × 10)

= 8000 kgm/s

The final momentum after collision will also be equal to 8000 kgm/s

Change in momentum = 8000 - 8000

Change in momentum = 0

5 0

3 years ago

Read 2 more answers

What force must be applied to move a 251 kg rock on a pavement like surface

Alchen [17]

If the rock is just sitting there and you want to SLIDE it, then you have to push it with a force of at least

(251 kg) x (9.8 m/s²) x (μ) =

(2,459 Newtons) x (the coefficient of static friction on that surface)

4 0

3 years ago

6. Mithra is an unknown planet that has two airless moons. A and B, in circular orbits around it.

Ainat [17]

The acceleration of gravity on Moon B is D) 0.25 m/s2

Explanation:

The data listed in the problem are not clear: find them in the table attached.

The acceleration of gravity on a planet is given by the following equation:

$g=\frac{GM}{R^2}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

M is the mass of the planet

R is the radius of the planet

Here we want to find the acceleration of gravity on the surface of Moon B, which has the following data:

$M=1.5\cdot 10^{20} kg$ (mass)

$R=2.0\cdot 10^5 m$ (radius)

And substituting into the equation, we find:

$g=\frac{(6.67\cdot 10^{-11})(1.5\cdot 10^{20})}{(2.0\cdot 10^5)^2}=0.25 m/s^2$

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

9. You are heating a piece of glass and now want to pick it up, You should

drek231 [11]

Answer:

c. use tongs

Explanation:

6 0

3 years ago

An infinite slab of charge of thickness 2z0 lies in the xy-plane between z=−z0 and z=+z0. The volume charge density rho(C/m3) is

Oksi-84 [34.3K]

Answer:

please read the answer below

Explanation:

To find the electric field you can consider the Gaussian law for a cylindrical surface inside the slab.

$\int E dA=EA_{G}=\frac{Q_{int}}{\epsilon_o}$

$Q_{int}=\rho V_{G}$

where Qint is the charge inside the Gaussian surface, AG is the area of the surface and rho is the charge density of the slab.

By using the formula for the volume of a cylinder you obtain:

$V_{G}=\pi r^2h$

where h is the height. If you assume that the slab is in the interval (-zo<z<z0) you can write VG:

$V_{G}=\pi r^2 z$

Finally, by replacing in the expression for E you get:

$E=\frac{Q_{int}}{\epsilon_o A_G}=\frac{Q_{int}}{\epsilon_o \pi r^2}\frac{z}{z}=\frac{\rho z}{\epsilon_o}$

$E=\rho z/\epsilon_o$

hence, for z>0 you obtain E=pz/eo > 0

for z<0 -> E=pz/eo < 0

7 0

3 years ago