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Zina [86]
4 years ago
6

A rocket is fired from rest from the ground (y = 0) at time t0 = 0 s. As the rocket is burning its fuel, it moves vertically upw

ard with a constant acceleration of 3.25 m/s2 . At t1 = 10.0 s, all the fuel has been used up and the rocket is in free fall. Air resistance can be neglected up until t3 = 17.5 s when the rocket deploys a parachute. At this time the rocket immediately reaches its terminal velocity; this means that the rocket is no longer accelerating.
1a. (10 points) Find the maximum height of the rocket with respect to the ground. How long after being launched does it take for the rocket to reach this height?
1b. (10 points) How long after being launched does it take for the rocket to return to the ground? What is the rocket’s displacement over this entire trip? What is the rocket’s average velocity over this entire trip? What is the rocket’s average speed over this entire trip? What is the rocket’s average acceleration over this entire trip? Be sure to include both magnitude and direction for any vector quantities.
1c. (10 points) What is the rocket’s displacement from ta = 12.0 s to tb = 20.0 secs? What is the rocket’s average velocity from ta to tb? What is the rocket’s average speed from ta to tb? What is the rocket’s average acceleration from ta to tb? Be sure to include both magnitude and direction for any vector quantities.
1d. (10 points) Sketch a y v. t plot for the motion of the rocket. Be sure to label the positions of the rocket at t0 = 0 s, t1 = 10.0 s, t2 (when the rocket is at its highest point), t3 = 17.5 s, and t4 (when the rocket returns to the ground).

Physics
1 answer:
elixir [45]4 years ago
6 0

Answer:

Explanation:

attached is the solution

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Consider the two-body situation at the right. A 3.50x103-kg crate (m1) rests on an inclined plane and is connected by a cable to
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Answer:a=2.42 m/s^2

Explanation:

Given

mass m_1=3.50\times 10^{3} kg

m_2=1.00\times 10^{3} kg

\theta(inclination)=30^{\circ}

\mu =0.21

Let T be the tension in the rope

From Diagram

m_1g\sin \theta -T-f_r=m_1a-----------------1

where f_r=friction\ force

f_r=\mu m_g\cos \theta

For block m_2

T=m_2a-----------2

From 1 & 2

m_1g\sin \theta -m_2a-\mu m_1g\cos \theta =m_1a

m_1g(\sin \theta -\mu \cos \theta )=(m_1+m_2)a

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4 0
4 years ago
Frictional force is____of area of contact​
Elis [28]

Answer:

The force due to friction is generally independent of the contact area between the two surfaces. This means that even if you have two heavy objects of the same mass, where one is half as long and twice as high as the other one, they still experience the same frictional force when you drag them over the ground.

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user100 [1]

Answer:

The average force has a magnitude 6524 N due north.

Explanation:

The average net force F = ma where m = mass of car = 1400 kg and a = acceleration.

a = (v - u)/t where u = initial velocity of car = 0 m/s (since it starts from rest)

v = final velocity of car = 27 m/s due north and t = time of motion = 5.8 s

a = (27 m/s - 0 m/s)/5.8 s = 27 m/s ÷ 5.8 s = 4.66 m/s

Since the direction of the velocity change is the direction of the acceleration, the acceleration is 4.66 m/s due north.

The average force, F = ma = 1400 kg × 4.66 m/s = 6524 N

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So the direction of the average force is due north

The average force has a magnitude 6524 N due north.

7 0
3 years ago
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