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marusya05 [52]
4 years ago
8

A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus

e the parachute is going down.) It comes to rest at 0 meters/second after 30 seconds. What's the average acceleration of the shuttle during this time span?
Physics
1 answer:
dexar [7]4 years ago
6 0

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

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On a horizontal, linear track lies a cart that has a fan attached to it. The mass of the cart plus fan is 364 g. The cart is pos
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Answer:

6.62s

Explanation:

Metric unit conversion:

364 g = 0.364 kg

789 g = 0.789 kg

Starting from rest, the cart takes 4.49 s to travel a distance of 1.43 m. We can use the following equation of motion to calculate the constant acceleration

s = a_1t_1^2/2

a_1 = \frac{2s}{t_1^2} = \frac{2*1.43}{4.49^2} = 0.142 m/s^2

Using Newton's 2nd law, we can calculate the force generated by the fan to push the 0.364 kg cart forward

F = a_1m_1 = 0.142*0.364 = 0.052 N

Now that more mass is added, the new acceleration of the 0.789 kg cart is

a_2 = F/m_2 = 0.052 / 0.789 = 0.065 m/s^2

We can reuse the same equation of motion to calculate the time it takes to travel 1.43 m from rest

s = a_2t_2^2/2

t_2^2 = 2s/a_2 = 2*1.43/0.065 = 43.7

t_2 = \sqrt{43.7} = 6.62s

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4 years ago
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a graph depicts force versus position. what represents the work done by the force over the given displacement?
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3 years ago
An oscillating dipole antenna 1.73 m long with a maximum 36.0 mV potential creates a 500 Hz electromagnetic wave. (a) What is th
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Answer:

E_0=0.021v/m

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From the question we are told that:

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