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charle [14.2K]
3 years ago
13

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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Let's say you go on a run and start to sweat because you are heating up.
Jobisdone [24]

Answer:

kinetic to thermal

Explanation:

4 0
3 years ago
Pls help answer embed <br>​
Savatey [412]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

Weight of the object = drag force

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

5 0
3 years ago
What is the relationship among
lawyer [7]
The bigger the object the greater the gravitational pull, so the farther away the big object is its gravitational force begins to decrease. Refer to the picture for more explanation.

4 0
3 years ago
The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic
stealth61 [152]

Answer:

Explanation:

1) The time of flight equation for projectile motion can be used here to find total time in air.

t = 2vsin∅ / g

where v is speed, Ф is launch angle

t = 2×4×sin 60 / 9.8

t = 0.71 seconds

2) Distance where it hit the ground is called as range and has the following standard equation

D = v² sin2Ф/g

D = 4²sin 2×60 / 9.8

D = 1.41m

3) Maximum elevation is maximum time reached

h = v² sin²Ф / 2g

h = 4²sin² 60 / 2*9.8

h = 0.61 m

3 0
3 years ago
A phone cord is 4.89 m long. The cord has a mass of 0.212 kg. A transverse wave pulse is produced by plucking one end of the tau
slega [8]

Answer:

Tension in Cord=174 N

Explanation:

Given Data

L (Phone Cord Length)=4.89 m

m (Cord Mass)=0.212 Kg

T (Time for four trips)=0.617 s

Tension=?

Solution

V=λ×f

V=\frac{8*4.89}{0.617}\\ V=63.4m/s

Sigma=\frac{mass}{length}\\ Sigma=\frac{0.212}{4.89}\\ Sigma=0.0433 \frac{kg}{m}

Wave Speed=\sqrt{\frac{Tension}{Sigma} }\\ \\V=\sqrt{\frac{T}{Sigma} }\\ V^{2}=\frac{T}{Sigma}\\  T=V^{2}*Sigma\\ T=(63.4)^{2}*(0.0433)\\ T=174 N

5 0
3 years ago
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