Ask question

Ask question

All categories

3 years ago

12

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

he same cart again traveling at 0.3 m/s collides with a different stationary object. This time the cart is at rest after the collision. In which collision is the impulse on the cart greater?
A. The first collision.

B. Cannot be determined without knowing the rebound speed of the first collision.

C. The second collision.

D. Cannot be determined without knowing the mass of the cart.

E. The impulses are the same.

1 answer:

Nady [450]3 years ago

6 0

The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse

You might be interested in

A county commissioner wants to reduce the amount of waste sent to the local landfill. Which of the following would result in the

Nookie1986 [14]

The recycling of glass, plastic, and metal would highly reduce the amount of the waste being sent to the local landfill. The large plastic container, fiberboards, etc, are removed by hand for this. This sorting makes it easy to recycle and reuse the plastic and fiber materials.

8 0

2 years ago

Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips

Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

$f = \frac{v}{\lambda}$

Then the relation between two different frequencies with same wavelength would be

$\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}$

$\frac{f'}{f} = \frac{v'}{v}$

The beat frequency heard when the two strings are sounded simultaneously is

$f_{beat} = f-f'$

$f_{beat} = f(1-\frac{f'}{f})$

$f_{beat} = f(1-\frac{v'}{v})$

We have the velocity of the transverse waves in stretched string as

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{200N}{\mu}}$

And,

$v' = \sqrt{\frac{196N}{\mu}}$

Therefore the relation between the two is,

$\frac{v'}{v} = \sqrt{\frac{192}{200}}$

$\frac{v'}{v} = \sqrt{0.96}$

Finally substituting this value at the frequency beat equation we have

$f_{beat} = 590(1-\sqrt{0-96})$

$f_{beat} = 11.92Hz$

Therefore the beats per second are 11.92Hz

4 0

2 years ago

In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single

FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is $2.27\times 10^{39}$

Explanation:

It is given that,

Distance between electron and proton, $r=4.53\ A=4.53\times 10^{-10}\ m$

Electric force is given by :

$F_e=k\dfrac{q_1q_2}{r^2}$

Gravitational force is given by :

$F_g=G\dfrac{m_1m_2}{r^2}$

Where

$m_1$ is mass of electron, $m_1=9.1\times 10^{-31}\ kg$

$m_2$ is mass of proton, $m_2=1.67\times 10^{-27}\ kg$

$q_1$ is charge on electron, $q_1=-1.6\times 10^{-19}\ kg$

$q_2$ is charge on proton, $q_2=1.6\times 10^{-19}\ kg$

$\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}$

$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}$

$\dfrac{F_e}{F_g}=2.27\times 10^{39}$

So, the ratio of electric force to the gravitational force is $2.27\times 10^{39}$ . Hence, this is the required solution.

3 0

3 years ago

Which device increases or decreases voltage as electricity is transmitted from power plants to homes and businesses?

Leona [35]

Answer:Transformer

Explanation:All the other devices generate their own electricity apart from power line and Transformer, power line just carries electricity. A transformer can be used to step up or step down voltages from a source.

4 0

3 years ago

Read 2 more answers

If the average distance between bumps on a road is about 10 m and the natural frequency of the suspension system in the car is a

Mnenie [13.5K]

When you hit a bump every 0.9 seconds.

3 0

3 years ago