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swat32
3 years ago
11

A 0.2 kg plastic cart and a 20 kg lead cart both roll without friction on a horizontal surface. Equal forces are used to push bo

th carts forward for a time of 1 s, starting from rest. Is the momentum of the plastic cart greater than, less than, or equal to the momentum of the lead cart?
Physics
1 answer:
garik1379 [7]3 years ago
8 0

Answer:

The same

Explanation:

The change in momentum of each cart is equal to the impulse given by the force applied on the cart:

\Delta p = I = F \Delta t

where

F is the force applied

\Delta t is the time during which the force is applied

The force applied to both carts (F) is the same, as well as the time (\Delta t). This means that the change in momentum, \Delta p, is the same for both carts. But both carts start from rest (momentum = 0): this means that at the end, both carts will have exactly same momentum.

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An object’s motion remains constant when acted upon by what?
igomit [66]

Answer:

An outside force

Explanation:

Newton's law an object in motion stays in motion an object at rest stays at rest unless acted on by an outside force.

6 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
Compare and contrast the general characteristics of a gas giant planet and one of the inner planets (such as Mercury, Earth, Ven
yKpoI14uk [10]
The gas planets usually have extremely high gravitational pulls, the surface isn't solid (since its a gas planet), and gas planets are larger than the inner planets. 
<span>Similarities- These planets all have moons and they both revolve around the sun (obviously).


Hope this helps.</span>
6 0
4 years ago
An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

n = 1

For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0
3 years ago
A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does n
Shkiper50 [21]

Answer:

Explanation:

velocity of sound in air at 20⁰C is 343 m /s

velocity of sound in water at 20⁰C is 1481 m /s

The wavelength of the sound is 2.86 m in the air so its frequency

= 343 / 2.86 = 119.93 .

This frequency of  119.93 will remain unchanged in water .

wavelength in water = velocity in water / frequency

= 1481 / 119.93

= 12. 35 m .

7 0
3 years ago
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