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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What will be the weight of a man of mass 90kg when he is on a planet with acceleration due to gravity of 15ms-2?

Feliz [49]

Answer:

1350N

Explanation:

Weight = Mass x Acceleration Due to Gravity

W=mg

W=90*15=1350N

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3 years ago

An object is lifted from the surface of aspherical planet to an altitude equal to the radius of the planet.As a result, what hap

serg [7]

An object is lifted from the surface of a spherical planet to an altitude equal to the radius of the planet.

As a result, the object's <em>mass remains the same</em>, and its <em>weight decreases</em> to 1/4 of whatever it is when the object is on the planet's surface.

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You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

$F=\frac{kQq}{r^2}$

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

$\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}$

And the force on the charge due to the charge on the north is,

$\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}$

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of $45^\circ$ from the north or from the north direction.