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liq [111]
2 years ago
13

A rocket is sitting on the launch pad. The engines ignite, and the rocket begins to rise straight upward, picking up speed as it

goes. At about 1000 m above the ground the engines shut down, but the rocket continues straight upward, losing speed as it goes. It reaches the top of its flight path and then falls back to earth. Ignoring air resistance, decide which one of the following statements is true.
a. Only part of the rocket's motion, from just after the engines shut down until just before it lands, is free-fall.
b. Only the rocket's motion, while the engines are firing, is free-fall.
c. Only part of the rocket's motion, from just after the engines shut down until it reaches the top of its flight path, is free-fall.
d. All of the rocket's motion, from the moment the engines ignite until just before the rocket lands, is free-fall.
e. Only the rocket's motion from the top of its flight path until just before landing is free-fall.
Physics
1 answer:
grandymaker [24]2 years ago
4 0

Answer:

e. Only the rocket's motion from the top of its flight path until just before landing is free-fall.

Explanation:

A free-fall  is a fall just under force of gravity. The rocket;s upward motion is result of engine push - even if it was shut down  - and rocket free of engine push effect when it reaches it's maximum height after shutting down of engine. Then rockets stops at it's maximum height for a moment and rtuens back as free fall with only force of gravitation pulling it back to ground with acceleration 'g'.

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What does the term equilibrium refer to?
snow_lady [41]
<h3><u>Answer</u>;</h3>

A) the resting position of the wave

<h3><u>Explanation</u>;</h3>
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  • <em><u>Equilibrium refers to a state of balance between opposing forces, it is a state of balance in which opposing forces cancel one another. </u></em>
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5 0
2 years ago
Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu
Anit [1.1K]
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 =  625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N


Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2


</span>
8 0
2 years ago
Read 2 more answers
Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They
denis-greek [22]

Answer:

  v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

         Em₀ = U = m g h

Final point. Lower

        Em_{f} = K = ½ m v²

        Em₀ = Em_{f}

        m g h = ½ m v²

         v² = 2gh

         v = √ 2gh

Let's calculate

         v = √ (2 9.8 150)

         v = 54.2 m / s

3 0
2 years ago
A student produces a power of p = 0.87 kw while pushing a block of mass m = 75 kg on an inclined surface making an angle of θ =
Paul [167]

When block is pushed upwards along the inclined plane

the net force applied on the block will be given as

F_{net} = mg sin\theta + \mu_k mg cos\theta

here we know that

m = 75 kg

\theta = 8.5 degree

\mu_k = 0.16

now plug in all values into this

F_{net} = 75\times 9.8 sin8.5 + 0.16 \times 75\times 9.8 cos8.5

F = 225 N

now for finding the power is given as

P = Fv

0.87 \times 10^3 = 225 \time v

v = \frac{870}{225} = 3.87 m/s

6 0
2 years ago
In a particular application involving airflow over a surface, the boundary layer temperature distribution may be approximated as
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Answer:

The Surface heat flux is -9205 W/m^2

Explanation:

 Explanation is in the following attachment    

8 0
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