Answer:
Kinetic energy = (1/2) (mass) (speed²)
Original KE = (1/2) (1430 kg) (7.5 m/s)² = 40,218.75 joules
Final KE = (1/2) (1430 kg) (11.0 m/s)² = 86,515 joules
Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules
Power = work/time = 46,296.25 joules / 9.3 sec = 4,978.1 watts .
Explanation:
Dont report my answer please
Answer:
If thermal energy is the motion energy of the particles of a substance, which has more thermal energy—the cup of hot tea or a spoonful of hot tea? It makes sense that the more particles of a substance you have, then the more thermal energy the substance has. The cup of hot tea would have more thermal energy, even if the temperature of the tea is the same in the cup and in the spoon. But which cools down the quickest (has the highest rate of thermal energy transfer)—the tea in the cup or the tea in the spoon? If I have fewer particles of the same substance, then the rate of thermal energy transfer is faster. The tea in the spoon would lose thermal energy more rapidly. So the amount of a substance you have is one factor that affects the rate of thermal energy transfer.
Explanation:
Answer:
Approximately 
(assuming that the acceleration due to gravity is
.)
Explanation:
Assuming that the weight on this 72-kg skydiver would be 
(points downwards.)
Air resistance is supposed to act in the opposite direction of the motion. Since this skydiver is moving downwards, the air resistance on the skydiver would point upwards.
Therefore, the net force on this skydiver should be the difference between the weight and the air resistance on the skydiver:
.
Apply Newton's Second Law of motion to find the acceleration of this skydiver:
.
Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
R= 12 ohm
2) circuit Q
R = 2.3 ohm
Volume = mass/density
Volume = 35000/1000
Volume = 35m^3
