How would the force of gravity between two objects change if the distance

Elena (60.0 kg) and Madison (65.0 kg) are ice-skating at the Rockefeller ice rink in New Yok city. Their friend Tanner sees Elen

Bas_tet [7]

1. +72.0 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

Taking "to the right" as positive direction, for Elena we have

m = 60.0 kg is the mass

v = +1.20 m/s is the velocity

So, Elena's momentum is

$p_e=(60.0 kg)(+1.20 m/s)=+72.0 kg m/s$

2. -162.5 kg m/s

Here Madison is moving in the opposite direction of Elena (to the left), so her velocity is

v = -2.50 m/s

while her mass is

m = 65.0 kg

Therefore, her momentum is

$p_m= (65.0 kg)(-2.50 m/s)=-162.5 kg m/s$

3. -90.5 kg m/s

The total momentum of Elena and Madison is equal to the algebraic sum of their momenta; taking into account the correct signs, we have:

$p=p_e + p_m = +72.0 kg m/s - 162.5 kg m/s =-90.5 kg m/s$

4. 0.72 m/s to the left

We can find the final speed of Elena and Madison by using the law of conservation of momentum. In fact, the final momentum must be equal to the initial momentum (before the collision).

The initial momentum is the one calculated at the previous step:

$p_i = -90.5 kg m/s$

while the final momentum (after the collision) is given by

$p_f = (m_e + m_m) v$

where

$m_e$ is Elena's mass

$m_m$ is Madison's mass

v is their final velocity

According to the law of conservation of momentum,

$p_i = p_f\\p_i = (m_e + m_m) v$

So we can find v:

$v=\frac{p_i}{m_e + m_m}=\frac{-90.5 kg m/s}{60.0 kg+65.0 kg}=-0.72 m/s$

and the direction is to the left, since the sign is negative.

8 0

3 years ago

A car traveling with an initial velocity of 10 m/s accelerates at a constant rate of 2.2 m/s^2 for 2 seconds. What distance does

mel-nik [20]

2.5 better than the first one more game to go hard but I 46 the bucks

5 0

3 years ago

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Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil

tatyana61 [14]

Answer:

The time taken by missile's clock is $4.6\times 10^{6} s$

Solution:

As per the question:

Speed of the missile, $v_{m = 6.5\times 10^{3}} m/s$

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

$T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}$ (1)

Using binomial theorem in the above eqn:

We know that:

$(1 + x)^{a} = 1 + ax$

Thus eqn (1) becomes:

$1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}$

$T = \frac{2c^{2}}{v_{m}^{2}}$

Now, putting appropriate values in the above eqn:

$T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}$

$T = 4.6\times 10^{6} s$

4 0

3 years ago

At what condition does a body becomes weightless at the equator?

ArbitrLikvidat [17]

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

4 0

2 years ago

When monochromatic light illuminates a grating with 7000 lines per centimeter, its second order maximum is at 62.4°. what is the

zhannawk [14.2K]

When red light illuminates a grating with 7000 lines per centimeter, its second maximum is at 62.4°. What is the wavelength of this light?

ans: 633nm

6 0

3 years ago

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