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Arlecino [84]
3 years ago
6

Which type of energy is water held back by a dam?

Physics
2 answers:
stepladder [879]3 years ago
8 0

Answer:    hydroelectric dams

Explanation: Enormous energy is stored in water held back by hydroelectric dams. The energy is transformed into (a form of kinetic energy) as it falls across the dam. The falling water strikes the blades of a turbine and makes them spin, and the then turns the shaft of a generator.

Setler [38]3 years ago
3 0

The answer is hydroelectric dams because Enormous energy is stored in water held back by hydroelectric dams. The energy is transformed into (a form of kinetic energy) as it falls across the dam.

Explanation: Please mark me as brainliest!

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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
  • Let's draw the free body diagram of the system using the given data.
  • From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
  • For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

                           N_x=86.62N

  • We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

                           N_y=F_V=mg-Tsin59\\

  • To find Ny, we need to find the tension T.
  • For this, we can equate the net horizontal force.

                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

  • Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

                    N_y= (40*9.8)-(169.8*sin59)=246.4N

  • Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

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in the figure shown, if the mass of the block is 4kg and the coefficient of static friction is 0.5 and the coefficient of kineti
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Explanation:

So since its saying whats the minimum F to move the block, we will use static friction (0.5).

We will use the equation for force of friction, which is Ff = uFn

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