The answer is : We’ll see the bell move, but we won’t hear it ring. This is because light can travel through vacuum but sound cannot. Sound waves are vibrations of particles in any media, so sound requires a medium to travel, and it cannot travel in a vacuum as there is no particles to vibrate.
Answer:
Explanation:
Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.
To objective is to find the:
(i) required heat exchanger area.
(ii) flow rate to be maintained in the evaporator.
Given that:
water temperature = 300 K
At a reasonable depth, the water is cold and its temperature = 280 K
The power output W = 2 MW
Efficiency
= 3%
where;



However, from the evaporator, the heat transfer Q can be determined by using the formula:
Q = UA(L MTD)
where;

Also;




LMTD = 4.97
Thus, the required heat exchanger area A is calculated by using the formula:

where;
U = overall heat coefficient given as 1200 W/m².K

The mass flow rate:

Explanation:
Displacement = 5 km
A.
Converting km/h to m/s,
10 km/h * 1000 m/1 km * 1 h/3600 s
= 25/9 m/s
Remember,
700 watt = 700 J/s
Velocity = displacement/time
Time = 5000/(25/9)
= 1800 s
Energy = power * time
= 700 * 1800
= 1,260,000
= 1260 kJ
B.
Converting km/h to m/s,
3 km/h * 1000 m/1 km * 1 h/3600 s
= 5/6 m/s
290 watt = 290 J/s
Velocity = displacement/time
Time = 5000/(5/6)
= 6000 s
Energy = power * time
= 290 * 6000
= 1,740,000
= 1740 kJ
C.
Walking burns more energy; 1,740,000 joules. It burns more because you walk for a greater period of time.
C.) In this, number of Hydrogen atoms is 4