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sergeinik [125]

3 years ago

8

Which of these statements are true about hydroelectric power? Check all that are true. It is the most widely used renewable ener

gy source in the world. It never disrupts ecosystems. It is not always available. A hydroelectric dam controls the force of moving water to generate electricity.

1 answer:

IRINA_888 [86]3 years ago

7 0

Answer:

the last one is triu!!!!!!

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How is the Sun related to nuclear electromagnetic and heat energy

Snezhnost [94]

-- The source of most of the energy that radiates from the sun is nuclear energy.

-- Most of the energy that radiates from the sun is electromagnetic energy.

-- Heat energy is part of the electromagnetic energy that radiates from the sun.
Other parts include radio, microwave, visible light, ultraviolet, and X-ray energy.

5 0

3 years ago

During an experiment, a scientist places a heat lamp above a bowl of water and uses the lamp to heat up the water. How does heat

alexandr402 [8]

Answer:

radiation

Explanation:

6 0

2 years ago

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Two scientists interpret the same set of data differently. How would the scientific community deal with this problem?

alexdok [17]

The correct answer would be D. A new experiment would be needed to be done in order to test the conclusions. In science there is no authority, data is the only thing that matters. So if we have two different conclusions from the same date the only solution is to perform more tests and more experiments to see what is correct.

7 0

3 years ago

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a

zepelin [54]

Answer:

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.

Explanation:

Given Values:

L = 50 cm = 0.5 m

H = 170 j/s

To find the diameter of the rod, we have to find the area of the rod using the following formula.

Here Tc = 100.0° C

k = 50.2

H = k × A × $\frac{[T_{H -}T_{C} ] }{L}$

Solving for A

A = $\frac{H * L }{k * [ T_{H}- T_{C} ] }$

A = $\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}$

A = $\frac{85}{12550}$ = 6.77 × $10^{-3}$ m²

Now Area of cylinder is :

A = $\frac{\pi }{4}$ d²

solving for d:

d = $\sqrt{\frac{4 * 0.00677 }{\pi } }$

d = 9.28 cm

5 0

3 years ago

A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$

where

m is the mass

a is the acceleration

In this problem, we have

$\sum F = 205.9 N$ is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2$

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

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

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

$a=1.37 m/s^2$

s = 350 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s$

4)

For this part of the problem, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

$a=1.37 m/s^2$

Solving for t, we find

$t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s$

Learn more about Newton's laws of motion and accelerated motion:

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6 0

3 years ago