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Zina [86]
3 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]3 years ago
6 0

Answer:

Explanation:

attached is the solution

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marishachu [46]

Answer:

The box will experience an acceleration.

Explanation:

Here, 2 N and 3 N forces are acting opposite to each other. In this case, the net force experience by the box would be (3-2)N = 1 N towards right. Since acceleration is directly proportional to the net force, therefore the box will experience an acceleration.

6 0
3 years ago
Which is NOT an example of evidence that shows a chemical reaction occurred?
Wittaler [7]
Change in color
Change in color is a physical reaction
4 0
3 years ago
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti
uysha [10]

Answer:The sled slides 16.875m before rest.

Explanation:

a=\frac{F}{m} =\frac{12N}{20kg}

a=0.6 m/s²

Vf=0=Vi-a.t

t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg

d= Vi.t - \frac{a.t^{2}}{2}

d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m

3 0
3 years ago
Problem 13.175 A 1-kg block B is moving with a velocity v0 of magnitude as it hits the 0.5-kg sphere A, which is at rest and han
Svetradugi [14.3K]

Explanation:

Mass of the Block, Mb = 1.0 kg Initial Velocity of Block, Vb = 2.0 m/s

Mass of the Ball, Ma = 0.5 kg Coefficient of Kinetic Friction, μ = 0.6

Coefficient of Restitution, e = 0.8 Gravity = 9.81 m/s^2

Sum of the forces of the X-axis components and Y-axis components are:

∑ Fx = Ff = Mb × a

Equation for frictional force,

Ff = μ × N

∑ Fy = N - mb × g = 0

Note:

N = mb × g

Therefore, to solve for the acceleration, we have:

μ × Mb × g = Mb × a

a = μ × g

= 0.6 × 9.81

= 5.88 m/s^2

Therefore, ratio of velocities using the coefficient of restitution, e:

e = (Vb2 – Va1)/ (Va – Vb)

Note: Va = zero (initially at rest)

Va2 – Vb2 = 0.8 × ( 0 - 2 m/s)

Vb2 – Va2 = -1.6 m/s

Vb2 = a2 – 1.6

Using Conservation of Momentum Equation:

Ma × Va + Mb × Vb = Ma × Va2 + Mb × Vb2

0 + 1kg × 2 m/s = 0.5 kg × Va2 + 1 kg × Vb2

Vb2 = 2 - 0.5 × Va2

Substitute in Vb2,

1.5 × Va2 = 3.6

Va2 = 2.4 m/s

Vbs = 0.8 m/s

.

mgh = 0.5 × Mv2

h = v2/ 2g

h = 0.294 m

B.

Using equation of motion,

Vf^2 = Vo^2 + 2a × S

Given:

Vf = 0 m/s

a = 5.88 m/s^2

0 = 0.82 + 2(5.88) × ΔS

Δx = 0.0544 m

4 0
3 years ago
0. What is centripetal accleration?Derive relation for it​
NISA [10]

Answer:

Centripetal acceleration is defined as the property of the motion of an object, traversing a circular path. Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration. ... Centripetal means towards the center.

4 0
3 years ago
Read 2 more answers
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