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Anna11 [10]
3 years ago
9

A rock falls from the top of a cliff. What will happen to the rock's kinetic energy when it strikes the ground below?

Physics
2 answers:
Paha777 [63]3 years ago
7 0

Answer:

B

Explanation:

It creates an equal amount of p energy.

REMEMBER: ENERGY CAN NEVER BE CREATED OR DESTROYED.

Salsk061 [2.6K]3 years ago
6 0
B. It will create an equal amount potential energy. (Don’t know if I’m right)
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dexar [7]

Answer: There is only one Sun in the galaxy … that is the thing that rises in the morning and sets at night. However, there is a use of “sun” to signify any old star … nobody knows exactly there might be trillions out there

Explanation:

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3 years ago
How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?
4vir4ik [10]

Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

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5 0
3 years ago
A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju
denpristay [2]

Answer:

a) h=250\ m

b) \Delta h=0.0835\ m

Explanation:

Given:

  • upward acceleration of the helicopter, a=5\ m.s^{-2}
  • time after the takeoff after which the engine is shut off, t_a=10\ s

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

h=u.t+\frac{1}{2} a.t^2

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

h=0+0.5\times 5\times 10^2

h=250\ m

b)

  • time after which Austin Powers deploys parachute(time of free fall), t_f=7\ s
  • acceleration after deploying the parachute, a_p=2\ m.s^{-2}

<u>height fallen freely by Austin:</u>

h_f=u.t_f+\frac{1}{2} g.t_f^2

where:

u= initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

t_f= time of free fall

h_f=0+0.5\times 9.8\times 7^2

h_f=240.1\ m

<u>Velocity just before opening the parachute:</u>

v_f=u+g.t_f

v_f=0+9.8\times 7

v_f=68.6\ m.s^{-1}

<u>Time taken by the helicopter to fall:</u>

h=u.t_h+\frac{1}{2} g.t_h^2

where:

u= initial velocity of the helicopter just before it begins falling freely = 0

t_h= time taken by the helicopter to fall on ground

h= height from where it falls = 250 m

now,

250=0+0.5\times 9.8\times t_h^2

t_h=7.1429\ s

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

t'=t_h-t_f

t'=7.1428-7

t'=0.1428\ s

<u>Now the height fallen in the remaining time using parachute:</u>

h'=v_f.t'+\frac{1}{2} a_p.t'^2

h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2

h'=9.8165\ m

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

\Delta h=h-(h_f+h')

\Delta h=250-(240.1+9.8165)

\Delta h=0.0835\ m

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