3 years ago

Why does the sound of something moving away from you seem to change to lower and lower pitch

Physics

1 answer:

lesantik [10]3 years ago

3 0

The doppler effect.
You are welcome.

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Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Fa

Morgarella [4.7K]

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

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2 years ago

The weight of an apple at 4r from earth is of that on earth

iris [78.8K]

W = m.g = weight
g = Gme/Re**2 where G= universal gravitational constant , Re= radius of the earth
me= mass of the earth
therefore it weighs 16 times less

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2 years ago

Read 2 more answers

A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force act

MA_775_DIABLO [31]

Answer:

cos 0 = 1.

Fs = 7×8 = 56 J

Explanation:

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3 years ago

A person must pull a rope 20 meters with a force of 200 N what is the work input?

tatiyna

Work done = force * distance
work done = 200 * 20
work done = 4000J

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3 years ago

A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?

stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, $m_1 = 300\ g = 0.3\ kg$

Initial speed of the cart, $u_1=1.2\ m/s$

Mass of the larger cart, $m_2 = 2\ kg$

Initial speed of the larger cart, $u_2=0$

After the collision,

Final speed of the smaller cart, $v_1=-0.81\ m/s$ (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let $v_2$ is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_1v_1=m_2v_2$

$v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}$

$v_2=0.301\ m/s$

So, the speed of the large cart after collision is 0.301 m/s.

4 0

3 years ago