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3 years ago

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How does the total momentum of the docked vehicle and station compare to the momentum of each object before the

1 answer:

Xelga [282]3 years ago

5 0

The total momentum before the docking maneuver is and after the docking maneuver is

Explanation:

Linear momentum (generally just called momentum) is defined as mass in motion and is given by the following equation:

Where is the mass of the object and its velocity.

According to the conservation of momentum law:

"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision will be the same as the total momentum of these same two objects after the collision ".

This means, that although the momentum of each object may change after the collision, the total momentum of the system does not change.

Now, the docking of a space vehicle with the space station is an inelastic collision, which means both objects remain together after the collision.

Hence, the initial momentum is:

Where:

is the mass of the vehicle

is the velocity of th vehicle

is the mass of the space station

is the velocity of the space station

And the final momentum is:

Where:

is the velocity of the vehicle and space station docked

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How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl

Vinvika [58]

Answer:

B) $\frac{2}{3}$

Explanation:

The electric force between charges can be determined by;

F = $\frac{kq_{1} q_{2} }{r^{2} }$

Where: F is the force, k is the Coulomb's constant, $q_{1}$ is the value of the first charge, $q_{2}$ is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

$q_{1}$ = 2q

$q_{2}$ = $\frac{q}{3}$

So that,

F = $q_{1}$ $q_{2}$ x $\frac{k}{r^{2} }$

= 2q x $\frac{q}{3}$ x $\frac{k}{r^{2} }$

= $\frac{2q^{2} }{3}$ x $\frac{k}{r^{2} }$

= $\frac{2}{3}$ x $\frac{kq}{r^{2} }$

F = $\frac{2}{3}$ x $\frac{kq}{r^{2} }$

The electric force between the given charges would change by $\frac{2}{3}$ .

4 0

3 years ago

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

Solved, can someone check it over!

kakasveta [241]

Answer:

its good no need to change anything :))

4 0

2 years ago

there are 5 influencing factors that may affect participation rate in physical activity. can you name them?

Stella [2.4K]

Demographic Barriers, Occupation, Age, Obesity, <span>
Psychological Barriers</span>

6 0

3 years ago

A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons

shepuryov [24]

Answer:

$v=2.556m/s$

Explanation:

From the conservation of mechanical energy

$K_{E1}+U_1=K_{E2}+U_2$

$\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2$

$x_2=0.08m$

$v_1=0 m/s$

Solve to velocity v2

$m*v_2^2=k*x_1^2-k*x_2^2$

$v^2=\frac{k}{m}*(x_1^2-x_2^2)$

$v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)$

$v=\sqrt{6.54m^2/s^2}=2.556m/s$

4 0

3 years ago