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nevsk [136]
3 years ago
7

) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene

rgy. How much power in generated by the turbine?
Physics
1 answer:
Angelina_Jolie [31]3 years ago
3 0

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

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Deep in the interiors of the giant planets, water is still a liquid even though the temperatures are tens of thousands of degree
Nataliya [291]

Answer:

High pressure inside the giant planet

Explanation:

As we move in the interior of the giant planet, the pressure and temperature in the interior of the planet increases. Since, the giant planets have hardly any solid surface and thus they are mostly constituted of atmosphere.

Also, the gravitational forces keep even the lightest of the matter bound in it contributing to the large mass of the planet.

If we look at the order of the magnitude of the temperature of these giant planets than nothing should be able to stay in liquid form but as the depth of the planet increases with the increase in temperature, pressure also increases which keeps the particle of the matter in compressed form.

Thus even at such high order of magnitude water is still found in liquid state in the interior of the planet.

7 0
3 years ago
Nuclear energy is over ___ times stronger than the chemical bonds between the atoms
Kamila [148]

Answer:

1 millions times stronger

8 0
2 years ago
Which factor below does NOT affect how fast a solute dissolves in a solvent?
vodomira [7]
There is no factor on your list of choices that has any effect.
7 0
3 years ago
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Eve makes a simple series circuit using a battery and two light bulbs. Through how many paths will the current be able to travel
lbvjy [14]

Answer:

Explanation:

3 I'm not sure u have insta

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3 years ago
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A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1
Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain d=2.46\times10^{3}\ m

Height of islandh=1.80\times10^{3}\ m

Distance of the enemy ship from the mountain d'=6.10\times10^{2}\ m

Initial velocity v=2.55\times10^{2}\ m/s

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

v_{x}=v\cos\theta

Put the value into the formula

v_{x}=2.55\times10^{2}\cos74.9

v_{x}=66.42\ m/s

We need to calculate the vertical component of initial velocity

Using formula of vertical component

v_{y}=v\sin\theta

Put the value into the formula

v_{y}=2.55\times10^{2}\sin74.9

v_{y}=246.19\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{v_{x}}

t=\dfrac{2.46\times10^{3}}{66.42}

t=37.03\ sec

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

H= v_{y}t-\dfrac{1}{2}gt^2

Put the value in the equation

H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2

H=2397.4\ m

We need to calculate the distance close to the peak

Using formula of distance

H'=H-h

Put the value into the formula

H'=2397.4-1800

H'=597.4\ m

Hence, The distance close to the peak is 597.4 m.

6 0
3 years ago
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