How can a river be used to produce electricity?

Physics

1 answer:

marysya [2.9K]4 years ago

4 0

Answer:

Scientists at IIT Roorkee are testing a floating device that can generate electricity from flowing surface water of rivers and streams, paving the way for an alternative source of renewable energy.

Explanation:

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What do you know about muscular fitness activities?

Ulleksa [173]

Answer:

Examples of muscle-strengthening activities include:

lifting weights.

working with resistance bands.

heavy gardening, such as digging and shovelling.

climbing stairs.

hill walking.

cycling.

dance.

push-ups, sit-ups and squats.

Explanation:

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3 years ago

A sky-diver jumps from a stationary balloon. His initial downwards acceleration is 10m/s².

Tatiana [17]

Explanation:

a. The force acting down is gravity, on Earth gravity is 10 m/s^2. When the skydiver jump, the acceleration will start out as -10 m/s^2, but it will eventually equals the air resistance , which is called terminal velocity.

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2 years ago

Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end

ozzi

Answer:

The total work that the rope does to Mangnus is - 5780 Jules.

Explanation:

By definition, the work is defined as:

$W=F.d$

Where F and d are the force and the total displacement. Note that in the definition the product is a scalar product since F and d are both vectors.

Take into account that according to third Newton's law the force that the rope does to Magnus is opposite to the force that Magnus does to the rope, therefore the scalar product will be negative due the rope's force goes against to Magnus displacement.

For calculating the work, we take 2500 N as the value for the force and 2.312 meters as the value for the displacement:

$W=-2500 N * 2.312 m$

$W=-5780 Nm = -5780 J$

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3 years ago

Rank the deformations of the following rods in terms of the magnitude of the average normal strain: (a) The length of a 1-m-long

kipiarov [429]

To solve this problem we will consider the concepts related to the normal deformation on a surface, generated when the change in length is taken per unit of established length, that is, the division between the longitudinal fraction gained or lost, over the initial length. In general mode this normal deformation can be defined as

$\epsilon = \frac{\delta}{l} = \frac{l_0-l}{l}$