Answer: The magnitude of the current in the second wire 2.67A
Explanation:
Here is the complete question:
Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?
Explanation: Please see the attachments below
Answer:
Option B. 3.0×10¯¹¹ F.
Explanation:
The following data were obtained from the question:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:
Capacitance (C) = Charge (Q) / Potential difference (V)
C = Q/V
With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
C = Q/V
C = 3.0×10¯⁹ / 100
C = 3.0×10¯¹¹ F.
Therefore, the capacitance of the parallel plate capacitor is 3.0×10¯¹¹ F.
Answer:
Distance = 3.69 × 10^9 m
The distance from the probe to Earth is 3.69 × 10^9 m
Explanation:
Distance from the probe to the Earth can be derived using the simple motion formula;
Distance = speed × time .....1
Since a radio signal uses an electromagnetic wave to transfer signal, it has the same speed as the speed of light.
Speed of radio signal = speed of light = 3.0 × 10^8 m/s
time taken to reach the earth = 12.3 seconds
Substituting the values of speed and time into equation 1;
Distance = 3.0 × 10^8 m/s × 12.3 s
Distance = 36.9 × 10^8 m
Distance = 3.69 × 10^9 m
Note: all electromagnetic radiation have the same speed which is equal to 3.0 × 10^8 m/s
Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ = 
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ = 
V = V₀
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀
0.1V₀ = V₀
0.1V₀/V₀ = 
0.1 = 
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s