The potential energy is 490 because
1•9.8•50=490 hope it helps!
Answer:
The displacement in t = 0,
y (0) = - 0.18 m
Explanation:
Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N
v = √ T / μ
v = √20.48 N / 0.02 kg /m = 32 m/s
λ = v / f
λ = 32 m/s / 40 Hz = 0.8
K = 2 π / λ
K = 2π / 0.8 = 7.854
φ = X * 360 / λ
φ = 0.5 * 360 / 0.8 = 225 °
Using the model of y' displacement
y (t) = A* sin ( w * t - φ )
When t = 0
y (0) = 0.25 m *sin ( w*(0) - 225 )
y (0) = 0.25 * -0.707
y (0) = - 0.18 m
Answer:
N1 / N2 = 1.0016
Explanation:
The apparent weight of the student is the value that a balance would have this corresponds in this case to the normal of the student. Let's write Newton's second law in the lower and upper part of the loop
For lower rotating wheel
N1 - W = m a
a = v² / r
N1 = W + v² / r
The relationship between linear and angular velocity is
v = w r
As the wheel rotates at a constant speed, we can use angular kinematics
w = θ / t
θ = 3 rev (2π rad / 1rev) = 6π rad
w = 6π / 150
w = 0.04π rad / s = 0.1257 rad / sec
calculate
N1 = mg + w² r
N1 = 55 9.8 + 0.1257² 27
N1 = 539 + 0.4267
N1 = 539.426 N
Now we perform the same calculation for the top
-N2 - W = -m a
N2 = -W + ma
N2 = 539 - 0.4267
N2 = 538.5733 N
The relationship between the weight at the bottom and top is
N1 / N2 = 539.4267 / 538.5733
N1 / N2 = 1.0016
Answer:
Explanation:
Resonant frequency is 240
4π² x 240² = 1 / LC
230400π² = 1 / LC
Let the required frequency = n
inductive reactance = 2 πn L
capacitative reactance = 1 / 2 π n C
inductive reactance / capacitative reactance
= 4π² x n ² x LC = 5.68
4π² x n ² = 1 / LC x 5.68
= 230400π² x 5.68
4n ²= 230400 x 5.68
n ²= 57600 x 5.68
n ² = 327168
n = 572 approx