Answer:
15 watt
Explanation:
Power is the rate at which work is done.
This means you divide the work done with the amount of time used to perform the work.
The formula for Power is : P = W/t where;
W= work done in J = 45
t= time in seconds = 3 sec
P= 45/ 3 = 15 watt
Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts
Answer:
680 J
Explanation:
Mechanical energy = potential energy + kinetic energy
ME = PE + KE
ME = mgh + ½ mv²
ME = (77.1 kg) (9.8 m/s²) (0.90 m) + ½ (77.1 kg) (0 m/s)²
ME = 680 J
Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:
KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J
Part B:
Now you can use Hooke’s law to find the force:
F = kx
F = (5000)(0.2)
F = 1000 N
The magnitude of work done by the gas is 279 J and the sign is negative so W = -279 J as work is done by the system.
<u>Explanation:</u>
According to first law of thermodynamics, the change in internal energy of the system is equal to the sum of the heat energy added or released from the system with the work done on or by the system. If the heat energy is added to the system to perform a certain work, then the heat energy is taken as positive, while it will be negative when the heat energy is released from the system.
Similarly, in this case, the heat energy of 597 J is added to the system. So the heat energy will be positive, while the gas expansion occurs means work is done by the system.
ΔU = Q+W
Since ΔU is the change in internal energy which is given as 318 J and the heat energy added to the system is Q = 597 J.
Then the work done by the gas = ΔU - Q = 318 J - 597 J = - 279 J.
As the work is done by the system, so it will be denoted in negative sign and the magnitude of work done by the gas is 279 J.
