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3 years ago

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A movie stunt double is supposed to run across the top of a train (in the opposite direction that the train is moving) and just

barely jump off before reaching a tunnel, but after running the entire length of the train (starting from the front). If the train is moving at 50 km/hr, is 3 km long and the tunnel is 40 km away from the end (where the stunt double is going to jump from), how fast (in km/hr) will the stunt double need to run

1 answer:

Rus_ich [418]3 years ago

3 0

Answer:

3.49km/hr

Explanation:

This train is moving at 50km/hr

Distance = length of train + distance of tunnel

= 3 km + 40 km

= 43 km

Time taken by train to get to tunnel

Time = distance / velocity

Time = 43 km/50km perhour

= 0.86

How fast will stunt double have to run?

Velocity = 3km/0.86

= 3.49 km/hr

The stunt double will have to run as fast as 3.49 kilometer per hour

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Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

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$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

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(b). Block acceleration is given by

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$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

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a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

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3 years ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

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Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

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$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

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Answer:

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Hello there!

For this:

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3). Use the following formula to solve for speed!

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My goal is to make sure you understand the problem, which is why I won't be giving you the answer. It'll be more work now, but less work in the future! :)

Hope this helped!

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