Answer:
Induced emf in the loop is 0.02208 volt.
Induced current in the loop is 0.0368 A.
Explanation:
Given that,
Area of the single loop,
The initial value of uniform magnetic field, B = 3.8 T
The magnetic field is decreasing at a constant rate,
(a) The induced emf in the loop is given by the rate of change of magnetic flux.
(b) Resistance of the loop is 0.6 ohms. Let I is the current induced in the loop. Using Ohm's law :
Hence, this is the required solution.
Answer:
m = 4.44 [g]
Explanation:
This is a problem related to heat transfer and thermodynamics. The second law of thermodynamics tells us that heat goes in only one direction, from the highest temperature substance to the lowest temperature. In this case the heat goes from tea to ice. The heat transfer process could be defined as the heat rejection of one body will be equal to the Heat received by another body.
Where:
m = mass = 177[g] = 0.177[kg]
Cp = specific heat = 4186 [J/kg*C]
T_f = 29.1 [°C]
Ti = 31.1 [°C]
Q = 0.177*4186*(29.1 - 31.1)
Q = - 1481.8 [J]
Note: The negative sign means that the heat is rejected.
Recall that the heat rejected is equal to the heat transferred
As the ice is going through a phase change, the fusion latent heat should be used, i.e. when it passes from solid to liquid. The heat transfer for this process is calculated with the following expression.
Qin = m *hf
m = mass of the ice [kg]
hf = Fusion latent heat = 333700 [J/kg]
1481.8 = m * 333700
m = 0.00444 [kg] = 4.44 [g]
Answer: the friction force on the small block at t equals 1 second is 2N
Explanation:
Given the data in the question;
from the slope in the graph provided, we will get the acceleration of the slab
At t = 1 seconds
Slope = acceleration = ( 1 - 0) / ( 2 - 0 ) = 1/2 = 0.5 m/s²
Force = ma = 4 × 0.5 = 2 N
so by Newton's third law
Force on block will be same which is 2N
Therefore the friction force on the small block at t equals 1 second is 2N
