Answer : The correct option is, (D) A machine does 400 joules of work in 5 seconds.
Explanation :
Power : It is defined a the rate of doing work per unit time.
Formula used :
where,
P = power
w = work done
t = time
Now we have to determine the rate of power for the following options.
(A) A machine does 200 joules of work in 10 seconds.
(B) A machine does 400 joules of work in 10 seconds.
(C) A machine does 200 joules of work in 5 seconds.
(D) A machine does 400 joules of work in 5 seconds.
From this we conclude that, a machine does 400 joules of work in 5 seconds has the highest rate of power.
Hence, the correct option is, (D)
Helping verbs come before the main verb, the main verb in the sentence is repair therefor the helping verb would be could.
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.
Answer:
1. F = M x A
2. Force
3. 2nd Law: Force
4. a, b, c (in order)
5. 3rd Law: Action and Reaction
6. b, c, a (in order)
7. 1st Law: Inertia
Answer:
Control of air–fuel ratio
Oxygen sensors tell the ECU whether the engine is running rich (too much fuel or too little oxygen) or running lean (too much oxygen or too little fuel) as compared to ideal conditions (known as stoichiometric).
Explanation:
