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Arlecino [84]
3 years ago
12

A rocket blasts off. In 10.0 seconds it is at 10,000 ft, traveling at 3600 mph. Assuming the direction is up, calculate the acce

leration. (Hint: the rocket is not under constant acceleration). 5280 ft/s2 528 ft/s2 100 ft/s2 200. ft/s2
Physics
2 answers:
marissa [1.9K]3 years ago
7 0

       Acceleration  =  (change in speed)  /  (time for the change)

Change in speed = (ending speed) - (beginning speed)

                            =     (3,600 mi/hr) - ( 0 )

                            =       3,600 mi/hr

                            =     same as  1 mile/second.

Acceleration  =  (1 mi/sec) / (10 sec)

                      =     0.1 mi/sec² .

But 1 mile = 5,280 ft,
so
                       0.1 mi/s²  =  528 ft/s²          
mojhsa [17]3 years ago
7 0

Answer:

 0.1 mi/s²  =  528 ft/s²    

Explanation:

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During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±
Alla [95]

Answer:

P = 1 (14,045 ± 0.03 )  k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

            Δ (Pₓ / Py) =?

 Let's start by finding the momentum of each vehicle

car X

        Pₓ = m vₓ

        Pₓ = 2.34 2.5

        Pₓ = 5.85 kg m

car Y

        Py = 2,561 3.2

        Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

          ΔPₓ = m Δv + v Δm

          ΔPₓ = 2.34 0.01 + 2.561 0.01

          ΔPₓ = 0.05 kg m

         ΔP_{y} = m Δv + v Δm

         ΔP_{y} = 2,561 0.01+ 3.2 0.001

         ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

          P = Pₓ / P_{y}

          ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

          ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

          ΔP = 0.006 + 0.0026

          ΔP = 0.009 kg m

The result is

           P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s

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3 years ago
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A hockey pick sliding along a frictional surface strikes a box at rest, after the collision the two objects stick together and m
Hitman42 [59]

Answer:

Explanation:

Option a is correct  

If puck and pick constitute a system then the momentum of the system is conserved but not this may not be valid for the puck .

Option e is correct

If puck and pick is the system then momentum is conserved but because of the presence of friction, mechanical energy is not conserved.

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What is the kinetic energy of a 0.25kg ball rolling at a speed of 2.5m/s
Lisa [10]

Answer:

Explanation:

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v=speed

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