Answer:
The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Explanation:
Given;
radius of the circular loop of wire = 0.5 m
current in circular loop of wire = 2 A
strength of magnetic field in the wire = 0.3 T
τ = μ x Bsinθ
where;
τ is the magnitude of the magnetic torque
μ is the dipole moment of the magnetic field
θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90
μ = IA
where;
I is current in circular loop of wire
A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²
μ = 2 x 0.7885 = 1.571 A.m²
τ = μ x Bsinθ = 1.571 x 0.3 sin(90)
τ = 0.4713 J
Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
The speed of the ball is always zero and the acceleration is always -g when it reaches the top of its motion. This is because when the ball is free, only gravity acts on it which is always downwards, hence g is the net acceleration and it is always negative. However the velocity does not direction change instantly, negative acceleration first slows down the ball with a positive velocity, until that point the ball keeps moving up, then the ball velocity becomes zero just before changing direction and becoming negative after which the ball will now go down along gravity. Hence the ball velocity is zero at the top (neither going up nor down). Mathematically this can be seen as velocity is the integration of acceleration.
The best conclusion to draw based on the description would be: <span>A.The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
This phenomenon happened because </span><span>The electric field from a positive charge will points away from the charge while the electric field from a negative charge will points toward the charge</span>
Answer:
0.4 ohms.
Explanation:
From the circuit,
The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.
1/R' = 1/R1+1/R2
R' = (R1×R2)/(R1+R2)
R' = (2.4×1.2)/(2.4+1.2)
R' = 2.88/3.6
R' = 0.8 ohms.
Hence the current flowing through the circuit is
I = V'/R'................ Equation 1
Where V' = voltmeter reading
I = 6/0.8
I = 7.5 A
This is the same current that flows through the variable resistor.
Voltage drop across the variable resistor = 9-6 = 3 V
Therefore, the resistance of the variable resistor = 3/7.5
Resistance = 0.4 ohms.
Answer:
2805 °C
Explanation:
If the gas in the tank behaves as ideal gas at the start and end of the process. We can use the following equation:
The key issue is identify the quantities (P,T, V, n) in the initial and final state, particularly the quantities that change.
In the initial situation the gas have an initial volume 
, temperature 
, and pressure 
,.
And in the final situation the gas have different volume 
and temeperature 
, the same pressure ,, and the same number of moles 
,.
We can write the gas ideal equation for each state:
and 
, as the pressure are equals in both states we can write
solving for
(*)
We know = 935 °C, and that the (the complete volume of the tank) is the initial volume plus the part initially without gas which has a volume twice the size of the initial volume (read in the statement: the other side has a volume twice the size of the part containing the gas). So the final volume 
Replacing in (*)
