<span>The particles through which compressional waves travel move in the same direction as the wave. This may be observed by fixing one end of a large spring and then compressing and extending the other end. The wave travels from one end to the other and the spring's parts move in the same direction.</span>
Answer:
The generator produces electrical energy at a rate of 1378125000 J per second.
Explanation:
volume of water falling each second is 1250 
height through which it falls, h is 150 m
mass of 1
of water is 1000 kg
⇒mass of 1250
of water, m = 1250×1000 = 1250000 kg
acceleration due to gravity, g = 9.8 
in falling through 150 m in each second, by Work-Energy Theorem:
Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it
⇒KE = mgh
= 1250000×9.8×150 J
= 1837500000 J
Electrical Energy =
(KE)
=
×1837500000
= <u>1378125000 J per second</u>
Answer:
1- For the track B. The potential energy is the same for the two cars, but because of the slope of the track, the car B earn kinetic energy faster. The gravitation acceleration of the cars will be g•sinθ, and the angle of the track B will have a bigger value for sinθ
2- The conservation of energy applies because the roller coaster is a closed track. When a car climb the track, it earn GPE, which is given by mgh, when it get down in the track, it transform GPE in KE, which is given in 1/2mv².
3-
Position of car (m) GPE KE GPE + KE
top (30m) 60000 0 60000
bottom (0m) 0 60000 60000
halfway down (15m) 30000 30000 60000
three-quarters way down 15000 45000 60000
Answer : The volume of a sample of 4.00 mol of copper is 
Explanation :
First we have to calculate the mass of copper.


Now we have to calculate the volume of copper.
Formula used :

Now put all the given values in this formula, we get:


Conversion used :

Therefore, the volume of a sample of 4.00 mol of copper is 
A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is
43.8 N = <em>k</em> (0.155 m) ==> <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.