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olganol [36]
4 years ago
13

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

Physics
1 answer:
Xelga [282]4 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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If two variables have a non-related relationship and one of the variables is changed, how will the other variable change?
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3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 44 m in front of you. Your reaction tim
never [62]

Answer:

14.0 m

25.1 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled in the reaction time

Distance = Speed × Time

\text{Distance}=20\times 0.5=10\ m

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20^2}{2\times -10}\\\Rightarrow s=20\ m

Distance in which the car will stop is 10+20 = 30.0 m

So, the car will not hit the deer

Distance between the car and deer is 44-30 = 14.0 m

\text{Distance}=u\times 0.5=0.5u\ m

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u^2=v^2-2as\\\Rightarrow u^2=0^2-2\times -10\times (44-0.5u)\\\Rightarrow u^2=20(44-0.5u)\\\Rightarrow u^2=880-10u\\\Rightarrow u^2+10u-880=0

u=\frac{-10+\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}, u=\frac{-10-\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}\\\Rightarrow u=25.08, -35.08\ m/s

Maximum speed of the car by which it will not hit the deer is 25.1 m/s

5 0
3 years ago
Read 2 more answers
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