Answer:
5x10^-3
Explanation:
Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.
Hooke's Law can be represented as
<h3> F = kx, </h3>
<em>where F is the force </em>
<em> k is the spring constant</em>
<em> x is the extension of the material </em>
<em />
Plug values in the equation
Step 1 find the original extension
0.045 = (400)x
x = 1.125x 10^-4 m d
Step 2 find the new extension
0.045+2 = 400(x)
2.045 = 400x
x = 5.1125x10^-3
Step 3 subtract the new extension with original
Total extension of the spring = 5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
Answer:
F. 25.82 s
Explanation:
Given:
Δy = 90 m
v₀ = 0 m/s
a = 0.27 m/s²
Find: t
Δy = v₀ t + ½ at²
90 m = (0 m/s) t + ½ (0.27 m/s²) t²
t = 25.82 s
