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3 years ago

How can water be used for energy???

Physics

2 answers:

Alla [95]3 years ago

4 0

Answer:

Water can be used for energy by placing dams in tidal areas.

Explanation:

Hydraulic energy is the energy supplied by the movement of water, in all its forms: waterfalls, rivers, sea currents, tides, waves, etc. This movement can be used directly, for example with a water mill, or more commonly be converted, for example into electrical energy in a hydroelectric plant.

Hydraulic energy is in fact a kinetic energy linked to the movement of water as in ocean currents, rivers, tides, waves or the use of potential energy as in the case of waterfalls and dams.

stellarik [79]3 years ago

3 0

<span>by placing dams in tidal areas. ITS the only one that has anything to do with water</span>

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Will give brainlist!! Please help!!

Marrrta [24]

Answer:Electromagnetic Energy Example One

activity: cellphones

type of electromagnetic: radio waves

description: we all use our phones to make phone calls and to send a text!

Electromagnetic Energy Example two

activity: microwave

type of electromagnetic: microwave radiation

description: The microwave radiation is absorbed by water molecules in the food which converts to heat intern heats the food do to high levels of radiation being emitted into the food!

Explanation:

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5 0

2 years ago

A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T

slamgirl [31]

Answer:

$u=14.48m/s$

Explanation:

From the question we are told that:

Height of window $h=2m$

Height of window off the ground $h_g=7.5m$

Time to fall and drop $t=1.3s$

Generally the Newton's equation motion is mathematically given by

$s=ut+\frac{1}{2}at^2$

Where

$h=ut+\frac{1}{2}at^2$

$2=u1.3-\frac{1}{2}*9.8*1.3^2$

$2=u1.3-8.281$

$u=7.91m/s^2$

Generally the Newton's equation motion is mathematically given by

$2as=v^2-u^2$

Where

$-2gh_g=v^2-u^2$

$-2*9.8*7.5=(7.91)^2-u^2$

$-147=62.5681-u^2$

$u=\sqrt{209.5681}$

$u=14.48m/s$

Therefore the ball’s initial speed

$u=14.48m/s$

8 0

2 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

5 0

3 years ago

For a transverse wave, what is a wavefront?

vichka [17]

Explanation:

A wavefront is the long edge that moves, for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v. These are drawn at a time t later, so that they have moved a distance s = vt.

4 0

2 years ago

Read 2 more answers

If a refrigerator is a heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the

Serjik [45]

The first law of thermodynamics states the conservation of energy and heat where the total energy in an isolated system may be transformed into another, but never created or destroyed. If 314 J of energy was released to the room, then also 314 J of energy was also removed from food in that refrigerator assuming it is an isolated system. <span>
</span>

7 0

3 years ago