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OverLord2011 [107]
3 years ago
11

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level.

The engines then fire, and the rocket accelerates upward at 4.10 m/s2 until it reaches an altitude of 1190 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.) (a) For what time interval is the rocket in motion above the ground
Physics
1 answer:
Dimas [21]3 years ago
5 0

Answer:

The rocket is motion above the ground for 44.7 s.

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine fails:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 =  initial height

v0 = initial speed

t=  time

a = acceleration due to the engines

g = acceleration due to gravity

First, let´s calculate the time the rocket is being accelerated by the engines:

(The center of the framer of reference is located at the ground, y0 = 0).

y = y0 + v0 · t + 1/2 · a · t²

1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²

0 = 2.05  m/s² · t² + 79.6 m/s · t - 1190 m

Solving the quadratic equation:

t = 11.5 s  (the other value of t is discarded because it is negative).

At that time, the engines fail and the rocket starts to fall. The equation of the height will be:

y = y0 + v0 · t + 1/2 · g · t²

The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:

v = v0 + a · t

v = 79.6 m/s + 4.10 m/s² · 11.5 s

v = 126.8 m/s

Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²

Solving the quadratic equation:

t = 33.2 s

Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s

 

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Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co
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Explanation:

The computation of the time required is shown below:

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So, the decelerative produced

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3 years ago
Batman (mass = 96.1 kg) jumps straight down from a bridge into a boat (mass = 458 kg) in which a criminal is fleeing. The veloci
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Answer:

The velocity of the boat after the batman lands in it is +9.26 m/s

Explanation:

Applying the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Note: The collision between the Batman and the boat is an inelastic collision.

m'u'+mu = V(m+m').................... Equation 1

Where m' = mass of the Batman, u' = initial velcoity of the batman, m = mass of the boat, u = initial velocity of the boat, V = common velocity.

make V the subject of equation 1

V = (m'u'+mu)/(m+m')............... Equation 2

Given: m' = 96.1 kg, u' = 0 m/s, m = 458 kg, u = +11.3 m/s.

Substitute these values into equation 2

V = [(96.1×0)+(458×11.2)]/(96.1+458)

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3 years ago
What is the displacement of the car between t=1s and t=4s
tensa zangetsu [6.8K]

Answer:

Option C. 30 m

Explanation:

From the graph given in the question above,

At t = 1 s,

The displacement of the car is 10 m

At t = 4 s

The displacement of the car is 40 m

Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:

Displacement at t = 1 s (d1) = 10 m

Displacement at t= 4 s (d2) = 40

Displacement between t = 1 and t = 4 (ΔD) =?

ΔD = d2 – d1

ΔD = 40 – 10

ΔD = 30 m.

Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.

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Which of the following is equivalent to 2.5 meters
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B is the correct answer for sure bro
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