Answer:
a) E = 0.0048 Volts
b) dA/dt = - 0.002285 m²/s
Explanation:
Given:
Area, A = 0.020 m²
Rate of change of magnetic field, dB/dt = 0.24 T/s
a) The magnitude of the emf induced (E) is given as:
E= A × (dB/dt)
on substituting the values in the above equation, we get
E = (0.020 m²) × (0.24 T/s)
or
E = 0.0048 Volts
b) Now, The induced emf when both the area and the magnetic field is varying
we have
E = B(dA/dt) + A(dB/dt)
Now, for the given case induced emf is zero i.e E = 0 and magnetic field B = 2.1 T
thus,
0 = (2.1 T)(dA/dt) + (0.020 m2)(0.24 T/s)
dA/dt = - 0.002285 m²/s
Hence, the area should be decreased at the rate of 0.002285 m²/s
Answer:
The object takes approximately 1.180 seconds to complete one horizontal circle.
Explanation:
From statement we know that the object is experimenting an Uniform Circular Motion, in which acceleration (
), measured in meters per square second, is entirely centripetal and is expressed as:
(1)
Where:
- Period of rotation, measured in seconds.
- Radius of rotation, measured in meters.
If we know that 
and 
, then the time taken by the object to complete one revolution is:
The object takes approximately 1.180 seconds to complete one horizontal circle.
Answer:
The velocity of a falling object
Explanation:
The positive X axis is towards right and positive Y axis is towards up, so North direction is positive
A vector with less than 1 magnitude is not negative, because its magnitude may be in between 0 and 1 which is positive vector.
Any vector whose magnitude is greater than 1 is never be a negative vector.
The velocity of a falling object is towards bottom, that is towards negative Y axis. So that vector is negative.
Answer:
A. 0.044 kg
Explanation:
We need to subtract the sum of (beaker+water - empty beaker) which is 0.106 kg - 0.062 kg = 0.044 kg. The answer will not be written in Newton because this unit is used for force only and in this question w have to find the weight.
Hope it is enough.
Please mark me as brainliest.
What happens is the potential value of the conductor decreases due to the presence of second conductor
as the capacitance is given by C = q/v
the value of v deceases as v-v1
thus the new capacitance is = C' = q/v-v1 thus the lowering of v increases the capacitance
