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Alexandra [31]
3 years ago
6

What type of power plant burns material to make electricity?

Physics
2 answers:
const2013 [10]3 years ago
8 0
<h2>Answer</h2>

Option D - Coal

<u>Explanation</u>

A fossil fuel power plant is a thermal power station in which fossil fuels are burned to produce electricity. Coal is a fossil fuel which is formed from the buried remains of plants and animals. Burning coal releases the energy that the plant absorbed from the sun. This energy is used to generate steam which is used to move a turbine to produce electricity. In this way, coal is burned by the power plants to produce electricity. In most of the countries, such plants provide most of the electrical energy used.

Alecsey [184]3 years ago
6 0
<h2>Hello!</h2>

The answer is: D. Coal

<h2>Why?</h2>

Coal power plants burn coal in to get steam, the steam flows into a turbine which is coupled to an electrical generator.

Coal power plants work burning high amounts of coal into a boiler, generation a lot of steam under extreme pressures. The steam is obtained when the water is heated by the burning coal, then the steam is cooled, being transformed in liquid water again (due the condensation process) and it's sent back on a cyclical process.

Have a nice day!

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Find acceleration. Will give brainliest!
zalisa [80]

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

7 0
3 years ago
Read 2 more answers
A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
rjkz [21]

Answer:

a) F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

b) \mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

\Sigma F_{y}=0

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

-F_{y}-W+N=0

When solving for the normal force we get:

N=F_{y}+W

we know that

W=mg

and

F_{y}=Fsin \theta

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

f_{k}=\mu_{k}N

so we can find the kinetic friction by substituting for N, so we get:

f_{k}=\mu_{k}(F sin \theta +mg)

Now we can find the sum of forces in x:

\Sigma F_{x}=0

so after analyzing the diagram we can build our sum of forces to be:

-f+F_{x}=0

we know that:

F_{x}=Fcos \theta

so we can substitute the equations we already have in the sum of forces on x so we get:

-\mu_{k}(F sin \theta +mg)+Fcos \theta=0

so now we can solve for the force, we start by distributing \mu_{k} so we get:

-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0

we add \mu_{k}mg to both sides so we get:

-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg

Nos we factor F so we get:

F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg

and now we divide both sides of the equation into (cos \theta-\mu_{k} sin \theta) so we get:

F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

f_{s}=\mu_{s}(F sin \theta +mg)

and

F cos θ = f

when substituting one into the other we get:

F cos \theta=\mu_{s}(F sin \theta +mg)

which can be solved for the static friction coefficient so we get:

\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

which is the answer to part b.

3 0
3 years ago
Read 2 more answers
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Sati [7]

Since you already gave us the weight of the 2.5-kg box,
we don't even need to know what the distance is, just
as long as it doesn't change.

Look at the formula for the gravitational force:

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If 'G', 'm₁' (mass of the Earth), and 'R' (distance from the Earth's center)
don't change, then the Force is proportional to  m₂ ... mass of the box,
and you can write a simple proportion:

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Cross-multiply:  (6.1 N) (1 kg)  =  (F) (2.5 kg)

Divide each side by (2.5 kg):  F = (6.1N) x (1 kg) / (2.5 kg)  =  2.44 N .

5 0
4 years ago
What is the difference between tangential acceleration and centripetal acceleration?
Nuetrik [128]

Centripetal acceleration is directed along a radius so it may also be called the radial acceleration. If the speed is not constant, then there is also a tangential acceleration (at). The tangential acceleration is, indeed, tangent to the path of the particle's motion.

4 0
3 years ago
A car moved 120km to the north. what is its displacement?
inn [45]
120Km. They are the same
5 0
3 years ago
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