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2 years ago

Which of the following describes the renewability of utility-scale wind energy?

Physics

1 answer:

schepotkina [342]2 years ago

6 0

Unlimited, clean energy plant describes the renewability of utility-scale wind energy.

<h3>What is utility-scale wind energy?</h3>

Utility-scale wind turbines are installed in large, multi-turbine wind farms connected to the nation's transmission system. These turbines produce clean and unlimited energy.

So we can conclude that Unlimited, clean energy plant describes the renewability of utility-scale wind energy.

Learn more about energy here: brainly.com/question/13881533

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Help ME please, I'M BENG TIMED

mixer [17]

Answer: The answer is A. Chemical

Explanation:Your hands get warm by fire because chemical energy gets converted into heat energy. When the chemical bonds in the wood are released in the air which then mixes with oxygen and emit heat.

6 0

2 years ago

A small uniform disk and a small uniform sphere are released simultaneously at the top of a high inclined plane, and they roll d

barxatty [35]

Answer:

(D) the sphere

Explanation:

The bodies given are Disk and Solid sphere (uniform sphere)

Moment of inertia of the bodies are

I(disk) = $\frac{MR^2}{2}$

I(sphere) = $\frac{2MR^2}{5}$

Since the moment of inertia of sphere is less than that of disk, therefore sphere will reach the bottom first.

3 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

The slider of mass m is released from rest in position A and slides without friction along the vertical-plane guide shown. Deter

Anuta_ua [19.1K]

The value of normal force as the slider passes point B is

6 mg

The value of h when the normal force is zero

3R/2

<h3>How to solve for the normal force</h3>

The normal force is calculated using the work energy principle which is applied as below

K₁ + U₁ = K₂

k represents kinetic energy

U represents potential energy

the subscripts 1,2 , and 3 = a, b, and c

for 1 to 2

K₁ + W₁ = K₂

0 + mg(h + R) = 0.5mv²₂

g(h + R) = 0.5v²₂

v²₂ = 2g(1.5R + R)

v²₂ = 2g(2.5R)

v²₂ = 5gR

Using summation of forces at B

Normal force, N = ma + mg

N = m(a + g)

N = m(v²₂/R + g)

N = m(5gR/R + g)

N = 6mg

for 1 to 3

K₁ + W₁ = K₃ + W₃

0 + mgh = 0.5mv²₃ + mgR

gh = 0.5v²₃ + gR

0.5v²₃ = gh - gR

v²₃ = 2g(h - R)

at C

for normal force to be zero

ma = mg

v²₃/R = g

v²₃ = gR

and v²₃ = 2g(h - R)

gR = 2gh - 2gR

gR + 2gR = 2gh

3gR = 2gh

3R/2 = h

Learn more about normal force at:

brainly.com/question/20432136

#SPJ1

8 0

1 year ago

A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a

Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

$v_x = \frac{x}{t}$