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sashaice [31]
2 years ago
9

Hydropower stations are almost always attached to dams, holding a large amount of water. But we can't dam air in a similar way t

o hold a store
of air for wind turbines.

Discuss the advantage that dams give to hydropower compared to wind power.

HELP AND ILL LIST YOU AS BRAINLIEST
Physics
1 answer:
Alja [10]2 years ago
7 0

Answer:

Advantages of Hydroelectric Energy

1 Renewable

Hydroelectric energy is renewable. This means that we cannot use up. However, there’s only a limited number of suitable reservoirs where hydroelectric power plants can be built and even less places where such projects are profitable.

2 Green

Generating electricity with hydro energy is not polluting itself. The only pollution occurs during the construction of these massive power plants.

3 Reliable

Hydroelectricity is very reliable energy. There are very little fluctuations in terms of the electric power that is being by the plants, unless a different output is desired. Countries that have large resources of hydropower use hydroelectricity as a base load energy source. As long as there is water in the magazines electricity can be generated.

4 Flexible

As previously mentioned, adjusting water flow and output of electricity is easy. At times where power consumption is low, water flow is reduced and the magazine levels are being conserved for times when the power consumption is high.

5 Safe

Compared to among others fossil fuels and nuclear energy, hydroelectricity is much safer. There is no fuel involved (other than water that is).

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Why is the motion of an athlete moving along the circular path with Constant speed considered to be an accelerated motion?
Strike441 [17]

The speed is changing its direction all the time. There is an acceleration which changes the direction of the speed – that is called centripetal acceleration. Only uniform linear motions are considered to have no acceleration.

This is the general formula for acceleration

a = dv/dt

When calculating dv, you should keep in mind the change in the velocity vector’s direction. You can easily see in a graph that with dt tending to 0 (so the length of the arc covered is also tending to 0), the difference between vectors Vf and V0 has a direction which is perpendicular to velocity (the shorter the arc, the closest the angle is to 90 degrees).

There is a formula (which can be deducted from the previous formula) which allows you to calculate the acceleration:

a = v^2/r

Let’s talk about the units:

v is in m/s

r is in m

so v^2/r

is in (m/s)^2/m = (m^2/s^2)/m = m/s^2

which is the same unit as dv/dt:

dv/dt = (m/s)/s= m/s^2

5 0
3 years ago
A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is
malfutka [58]

Answer:  A) 6.38(10)^{6} m

Explanation:

The equation for the moment of inertia I of a sphere is:

I=\frac{2}{5}mr^{2} (1)

Where:

I=9.74(10)^{37}kg m^{2} is the moment of inertia of the planet (assumed with the shape of a sphere)

m=5.98(10)^{24}kg is the mass of the planet

r is the radius of the planet

Isolating r from (1):

r=\sqrt{\frac{5I}{2m}} (2)

Solving:

r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}} (3)

Finally:

r=6381149.077m \approx 6.38(10)^{6} m

Therefore, the correct option is A.

4 0
3 years ago
A piano tuner is using a 392 Hz tuning fork to tune the wire for a G-Natural note. She hears 4 beats per second. What are the tw
inysia [295]

A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies. Frequency beat is equal to,

f_{beat} =| f_2\pm f_1 |

The reference frequency in our case would be 392Hz, and since there is the possibility of the upper and lower range for the amount of beats per second that the two possible frequencies are heard would be

f_{beat} =|392+4|= 396Hz

f_{beat} =|392-4|=388Hz

Therefore the two possible frequencies the piano wire is vibrating at, would be 396Hz and 388Hz

5 0
3 years ago
The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou
777dan777 [17]

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height  is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

3 0
4 years ago
You run<br> completely around a 400m track in<br> 80s. What was your average velocity?
dalvyx [7]

Answer:

V=?

S=400m

t=80s

V=S/t

V=400/80

V=5m/sec

6 0
3 years ago
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