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zavuch27 [327]
3 years ago
5

A stone is dropped from a cliff and falls 9.44 meters. What is the speed of the stone when it reaches the ground?

Physics
1 answer:
Lunna [17]3 years ago
3 0

Answer:

option A

Explanation:

given,                                              

height of the drop of stone = 9.44 m

speed of the stone = ?                          

As the stone is dropped the energy of the stone will be conserved.

using conservation of energy.            

Potential energy = Kinetic energy    

m g h = \dfrac{1}{2} m v^2  

     v = \sqrt{2gh}                  

     v = \sqrt{2\times 9.8 \times 9.44}

     v = \sqrt{185.024}              

            v = 13.60 m/s                      

Hence, the correct answer is option A

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Answer:

Height h = 37.8 m

Explanation:

Given :

Velocity of car (v) = 98 km / h

Acceleration of gravity = 9.8 m/s²

Computation:

Acceleration of gravity = 9.8 m/s²

Acceleration of gravity = (98)(1,000 m / 3,600 s)

Acceleration of gravity = 27.22 m/s

By using law of conservation of energy ;

(1/2)mv² = mgh

h = v² / 2g

h = 27.22² / 2(9.8)

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You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

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  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
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Answer:25km/s

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How much momentum will a dumb-bell of mass 10 kg transfer
frosja888 [35]

We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:

Vf² = Vi² + 2ad

Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.

Given values:

Vi = 0m/s (dumbbell starts falling from rest)

a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)

d = 80×10⁻²m

Plug in the values and solve for Vf:

Vf² = 2(10)(80×10⁻²)

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Reject the negative root.

Vf = 4m/s

The momentum of the dumbbell is given by:

p = mv

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Given values:

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6 0
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