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Marianna [84]
3 years ago
7

What energy transformation occurs during the combustion of coal in a power plant?

Physics
2 answers:
Leviafan [203]3 years ago
5 0
C. chemical to thermal
inn [45]3 years ago
4 0
Hello there, the correct answer is:

B.
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delivery ladies is Shivam noted at the thundering sound of 6 second after the lightning was seen by him​
vitfil [10]

Answer:

this is due to difference in speeds of sound and light

Explanation:

light has a speed of 3×10^8 m/s and it is seen at once because it takes negligible time due to very high speed and short distance  that is why as soon as lightning occurs we can see it . since thundering sound travels with speed of sound which is about 330  to 340 m/s in air hence it takes some time as described in question as 6 seconds

8 0
3 years ago
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A ringing bell sends sound waves in all<br> directions<br> places<br> sides
Strike441 [17]

Answer:

direction

Explanation:

because particles surround the bell, so when the bell vibrates, it causes particles surrounding it to vibrate back and forth vigorously. as these particles vibrate they collide with the neighbouring particles, passing on the energy.

hope this is what you are asking, if not please report it so that someone else gets to try it.

4 0
3 years ago
A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w
steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

h = L - L\cos{31°} = L(1 - \cos{31°})

so its initial gravitational potential energy U_0 is

U = mgh = mgL(1 - \cos{31°})

\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})

\:\:\:\:\:=0.23\:\text{J}

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U

We know that the initial kinetic energy K_0, as well as its final gravitational potential energy U are zero so we can write the conservation law as

mgL(1 - \cos{31°}) = \frac{1}{2}mv^2

Note that the mass gets cancelled out and then we solve for the velocity v as

v = \sqrt{2gL(1 - \cos{31°})}

\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}

\:\:\:\:\:= 1.3\:\text{m/s}

5 0
3 years ago
Read 2 more answers
Why is tungsten used as a filament in a bulb?
Viktor [21]

Answer:

because of tungsten's high melting point

Explanation:

7 0
3 years ago
a particle moves along the x axis with an acceleration of a=18t, where a has units if m/s2. if the particle at time t=0 is at th
ludmilkaskok [199]

Answer:

 Position at t= 4 seconds is 144 m

Explanation:

 It is given that acceleration, a = 18 t, where t is the time.

 We know that Velocity, v = \int { a} \, dt

  Substituting value of a,

           Velocity, v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c

 We know that at t = 0, v = -12 m/s

         So, 9*0^2+c=-12\\ \\ c=-12m/s

So velocity, v = (9t^2-12)m/s

  We also know that displacement, x = \int { v} \, dt

     Substituting value of v,  

        Displacement, x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c

  We know that at t = 0, particle is at origin, x =0.

               So,  0=3*0^3-12*0+c\\ \\ c=0

   Displacement, x = 3t^3-12t

At t = 4 seconds

   x = 3*4^3-12*4=192--48=144m

Position at t= 4 seconds is 144 m  

4 0
3 years ago
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