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Wewaii [24]
3 years ago
6

in coal-fired power plants, coal is burned to produce ____________________, which is then used to turn a turbine.

Physics
2 answers:
crimeas [40]3 years ago
6 0
The answer is STEAM.
Colt1911 [192]3 years ago
6 0

Answer:

The answer is steam!!  

Explanation:

This has been the case since the beginning, the coal is burned in a broiler, the broiler then produces steam; the steam turns the turbines, and the turbines turn the generator which turns the steam to electricity

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Cam Newton can sprint 40 meters in 5.79 seconds! How fast can he run?<br> Show your work
Setler79 [48]

Speed=Distance/Time

Distance=40m,time=5.79seconds

S=40/5.79

=6.908m/s

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How many atoms of carbon would two molecules of glucose (C6H12O6) have?
Sauron [17]
Well according to the molecular formula of glucose, one molecule would have 6 carbon atoms, and thus 2 molecules of glucose would have 12 carbon atoms.

The correct response would be B. 12.
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Which snack is an excellent choice for protein, B vitamins, vitamin E, and healthful fat
igomit [66]
Fruits and frozen fruit bars is the correct answer.
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A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is supported by a horizontal surface. Neglecting frictio
xxTIMURxx [149]

The angle of the wedge is 30°.

Answer:

5.88 ft/s

Explanation:

a) The block will slide down due to it's weight.

initial velocity u= 0

final velocity, v

acceleration, a = g sin 30° = 32 ft/s²× sin 30° = 16 ft/s²

Sliding displacement, s = 3ft

Use third equation of motion:

v^2-u^2 = 2as

substitute the values and solve for v

v^2-0 = 2\times 16 \times 3 =96 ft^2/s^2\\v = 9.8 ft/s

b) Use conservation of momentum:

Initial momentum of the system  = 0

final momentum = (15) ( 9.8)+ (25)(v')

v' = 5.88 ft/s

3 0
3 years ago
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
3 years ago
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