Answer:
22.2 W
Explanation:
First of all, we calculate the work done by moving the wagon, using the formula:
where
F = 20 N is the magnitude of the force
d = 1000 m is the displacement of the wagon
is the angle between the direction of the force and of the displacement (assuming the force is applied in the direction of motion)
Substituting, we find
Now we can find the power generated, which is equal to the ratio between the work done and the time taken:
where
W = 20,000 J
t = 15 min = 900 s
Substituting,
And the same value in Joules/second (remember that 1 Watt = 1 Joule/second)
The answer is False, pls. mark me the brainliest if I’m right. THX
Answer:
Explanation:
Magnitude of force per unit length of wire on each of wires
= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .
Putting the values ,
force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )
= .67 i² x 10⁻⁴
force on 3 m length
= 3 x .67 x 10⁻⁴ i²
Given ,
8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²
i² = 3.98 x 10⁻²
i = 1.995 x 10⁻¹
= .1995
= 0.2 A approx .
2 i = .4 A Ans .
The charge accumulated in 3.25 μF capacitor is 178.75 μC.
Answer:
Explanation:
In parallel connection, the voltage drop across any passive devices like capacitor or resistor will be constant. So the current flow will be varying in case of parallel connections of capacitors or resistors.
As the capacitance is the measure of amount of charge or current generated for a given amount of voltage, it is directly proportional to the charge or current and inversely proportional to voltage.
C = Q/V
Here the charge accumulated in a capacitor of capacitance 3.25 microfarad need to be determined which is in parallel connection with another capacitor. So the voltage through both the capacitor will be equal to the voltage of the battery which is stated as 55 V.
3.25×
= Q/55
Q = 3.25 * 55 = 178.75 μC
So the charge accumulated in 3.25 μF capacitor is 178.75 μC.
