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4 years ago

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:

hope it helps my dear friend.

You might be interested in

What aspect of motion can you conclude is common among freely falling objects ?

Liula [17]

Answer:

Free-fall is defined as the movement where the only force acting on an object is the gravitational force.

By the second Newton's law, we have that:

F = m*a

Where F = Force, m = mass, a = acceleration.

We can write this as:

a = F/m

And the gravitational force can be written as:

F = (G*M/r^2)*m

Where G is the gravitational constant, M is the mass of the Earth in this case, and r is the distance between both objects (the center of the Earth and the free-falling object)

As the radius of the Earth is really big, the term inside the parentheses is almost constant in the region of interest, then we can write:

G*M/r^2 ≈ g

And the gravitational force is:

F = g*m

And by the second Newton's law we had:

a = F/m = (g*m)/m = g

a = g

Then the acceleration does not depend on the mass of the object.

Then the thing that is common among the free-falling objects is the vertical acceleration.

6 0

3 years ago

What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation?.

Ber [7]

Answer:

It has no effect on the amplitude.

Explanation:

When the sandbag is dropped, then the cart is at its maximum speed. Dropping the sand bag does not affect the speed instantly, this is because the energy remains within the system after the bag as been dropped. The cart will always return to its equilibrium point with the same amount of kinetic energy, as a result the same maximum speed is maintained.

5 0

3 years ago

An object is located in air, 25 cm from the vertex of the concave surface of a block of glass (as viewed from the air side of th

Marizza181 [45]

Your question is missing one part as "magnification"

i have completed the missing part below

Answer:

a. $d_{i}=-0.0566cm$

b. $M=441.69$

Explanation:

For this type of numerical we will use the following formulas

$\frac{n1}{d_{o} }+\frac{n_{2} }{d_{i} }=\frac{n_{2}-n_{1} }{R}$ ......... Eq1

where,

$n_{1}$ $=$ refractive index of the medium surrounding refracting surface/object

i.e. $n_{1}$ = $n_{air}$

$n_{2}$ $=$ refractive index of the refracting surface/object

i.e. $n_{2}=n_{glass}$

$d_{0}=$ distance of object from the vertex of the refracting surface

$d_{i}=$ distance of image from the vertex of the refracting surface

$R=$ radius of curvature of the refracting surface

$M=\frac{d_{0} }{d_{i} }$ ........... Eq2

where,

$M=$ magnification

Convention:

$R>0\for\ the\ convex\ refractive\ surface\ of\ curvature\\\ R$

Given:

$n_{1}$ = $n_{air}$ $=1.0$

$n_{2}=n_{glass}$ $=1.5$

$d_{0}=$ $25cm$

$R=-11$ $cm$ because refraction surface is concave

Required:

a. $d_{i}=$ $?$

b. $M=?$

Solution:

a. putting values in eq1, we get

$\frac{1.0}{25}+\frac{1.5}{d_{i} }=\frac{1.5-1.0}{-11}$

$\frac{1.5}{d_{i} }=-0.045-0.040$

$d_{i}=(-0.085)(\frac{1}{1.5} )$

$d_{i}=-0.0566$ cm

b. $M=\frac{25}{-0.05665}$

$M=441.69$

5 0

4 years ago

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

Dmitriy789 [7]

Answer:

d = 2.54 [m]

Explanation:

Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

$E_{k1}+W=E_{k2}\\$