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pychu [463]
3 years ago
15

An object falls from a height of 12m. The object's potential energy at the top is 50J. Where is the potential energy of the obje

ct
25J
Physics
1 answer:
alex41 [277]3 years ago
5 0

Answer:

the height of the is 6 m.

Explanation:

Given;

potential energy of the object, P.E = 50 J

Height of fall of the object, H = 12 m

Determine the mass of the object, m

P.E = mgh

m = (P.E) / (gh)

m = (50) / (9.8 x 12)

m = 0.425 kg

Let the height of the object at 25 J = h₂

mgh₂ = 25

h₂ = (25) / (mg)

h₂ = (25) / (0.425 x 9.8)

h₂ = 6 m

Therefore, the height of the object at 25 J is 6 m.

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Setler79 [48]

Answer:

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3 0
3 years ago
A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and
Alexxandr [17]

Answer:

a

  F  =326.7 \ N

b

  M  = 6

Explanation:

From the question we are told that

          The mass of the rock is  m_r  =  200 \ kg

          The  length of the small object from the rock is  d  =  2 \ m

          The  length of the small object from the branch l  =  12 \ m

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      W =  m_r *  g

substituting values

     W =   200 *  9.8

     W =   1960 \ N

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         \sum  M_f  =  F * cos \theta *  l  -  W cos\theta  *  d =  0

Here  \theta is very small so  cos\theta  *  l  =  l

                               and  cos\theta  *  d  =  d

Hence

       F *   l  -  W  * d =  0

=>    F  = \frac{W * d}{l}

substituting values

        F  = \frac{1960 *  2}{12}

       F  =326.7 \ N

The  mechanical advantage is mathematically evaluated as

          M  = \frac{W}{F}

substituting values

        M  = \frac{1960}{326.7}

       M  = 6

6 0
3 years ago
A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
ankoles [38]
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

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3 years ago
WILL MARK BRAINLIEST!!!!!!!!!!!!!!!!!!
stepan [7]
*FRICTIONAL FORCE* in the opposite direction of the way Bobby is pushing.
Friction is a force which varies but it is always opposing the direction of motion.

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An applied force is literally the force that is applied to an object.

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It is the weight of the object multiplied by the force of gravity. If the TV weighed 100kg, acceleration is always 9.81 m/s^2, so the weight force would be 981 N.

*NORMAL FORCE* is the force which is holding the TV above ground. The ground supplies a force upward against the TV.
Normal force is just the force that prevents the TV from falling through the ground. We don't normally realize it in our everyday life, but the floor must hold everything up because gravity is always "pushing" against it.
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3 years ago
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mariarad [96]
Artery im pretty sure 
7 0
3 years ago
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