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liq [111]
1 year 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]1 year 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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Two children stand on a platform at the top of a curving slide next to a backyard swimming pool. At the same moment the smaller
victus00 [196]

1.

Answer:

a) It is less

Explanation:

By energy conservation we can say that initial potential energy of both child must be equal to the final kinetic energy of the two child.

Since initially they are at same height so we will say that initial potential energy will be given as

mgH and MgH

so the child with greater mass has more energy and hence smaller child will reach with smaller kinetic energy

2.

Answer:

b. The two speeds are equal.

Explanation:

As we know by mechanical energy conservation law we have

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

since both child starts at same height so here they both will reach the bottom at same speed

3.

Answer:

c. The two accelerations are equal

Explanation:

Since we know that average acceleration of the motion is given as

a = \frac{v_f - v_i}{\Delta t}

since here initial and final speeds are same so they both must have same average acceleration here.

5 0
2 years ago
A sleeping 68 kg man has a metabolic power of 79 w .
Lesechka [4]
 <span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>
3 0
2 years ago
The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is
Taya2010 [7]

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

            W = ΔEm

Where the work is

             W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

           Em₀ = K = ½ m v²

Final. When the system is at rest

            Em_{f} = K_{e1}b +K_{e2} = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

            ∑ F = F - fr

Axis y

           N- W = 0

           N = W

The friction force has the equation

          fr = μ N

          fr = μ W

  The job

         W = (F – μ W) x

We substitute in the equation

            (F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

           ½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values ​​and solve

           ½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

         x² 30 + 14.4 x - 1,125 = 0

        x² + 0.48 x - 0.0375 = 0

           

We solve the second degree equation

        x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

        x = [-0.48 ± 0.617] / 2

        x₁ = 0.0685 m

        x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

4 0
2 years ago
Janice is unsure about her future career path. She has grown up on her family farm, but she is also interested in medicine. Jani
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Answer:

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3 0
2 years ago
Read 2 more answers
What is the time spent by the alpha particle in the magnetic field?
Ber [7]

the alpha-particle goes around a quarter of the circle, so the time it takes would be t=0.25×2.61×10−6s=6.5×10−7s.

I hope this help
7 0
2 years ago
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