Answer:
The current is changing at the rate of 0.20 A/s
Explanation:
Given;
inductance of the inductor, L = 5.0-H
current in the inductor, I = 3.0 A
Energy stored in the inductor at the given instant, E = 3.0 J/s
The energy stored in inductor is given as;
E = ¹/₂LI²
E = ¹/₂(5)(3)²
E = 22.5 J/s
This energy is increased by 3.0 J/s
E = 22.5 J/s + 3.0 J/s = 25.5 J/s
Determine the new current at this given energy;
25.5 = ¹/₂LI²
25.5 = ¹/₂(5)(I²)
25.5 = 2.5I²
I² = 25.5 / 2.5
I² = 10.2
I = √10.2
I = 3.194 A/s
The rate at which the current is changing is the difference between the final current and the initial current in the inductor.
= 3.194 A/s - 3.0 A/s
= 0.194 A/s
≅0.20 A/s
Therefore, the current is changing at the rate of 0.20 A/s.
Answer:
3.5 cm
Explanation:
mass, m = 50 kg
diameter = 1 mm
radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m
L = 11.2 m
Y = 2 x 10^11 Pa
Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3
= 7.85 x 10^-7 m^2
Let the wire is stretch by ΔL.
The formula for Young's modulus is given by
ΔL = 0.035 m = 3.5 cm
Thus, the length of the wire stretch by 3.5 cm.
Answer:
0.6A
Explanation:
Area of loop =200cm2 =200 x10 ∧-4m∧2 Change in Magnetic field (B)= 25mT -10mT =15mT time =5ms
From Faraday' s law of induction EMF(E)= change in magnetic field/time
E= 15mT/5ms
Note, that one weber per second is equivalent to one volt.
= 3V
from Ohm's law I =E/R
=3/5 =0.6A
Answer:
0
Explanation:
Assuming your graph and question match the attachment, the average speed is 0. The bug ends up where it started, so its displacement is zero.
average speed = displacement/time = 0/(8 s)
average speed = 0
