To develop this problem it is necessary to apply the concepts related to Sound Intensity.
By definition the intensity is given by the equation

Where,
I = Intensity of Sound
= Intensity of Reference
At this case we have that 15 engines produces 15 times the reference intensity, that is

And the total mutual intensity is 100 dB, so we should




Therefore each one of these engines produce D. 88dB.
:<span> </span><span>30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s
At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg)
So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road ..
Fn = mg - mv²/R
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards)
(b)
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero.
ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)²
►v = √198 = 14.0 m/s</span>
D:the electrons from being attracted to the grid instead of the anode
Answer : The rate of heat transfer to the water is, 37.92 kJ/min
Explanation : Given,
Time = 10 min
Mass of water = 200 g
Latent heat of fusion of water = 334 J/g
Latent heat of vaporization of water = 2230 J/g
Now we have to calculate the rate of heat transfer to the water.

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


Thus, the rate of heat transfer to the water is, 37.92 kJ/min