Answer:
v_f = 1.05 m/s
Explanation:
From conservation of energy;
E_f = E_i
Thus,
(1/2)m(v_f)² + (1/2)I(ω_f)² + m•g•h_f + (1/2)k•(x_f)² = (1/2)m(v_i)² + (1/2)I(ω_i)² + m•g•h_i + (1/2)k•(x_i)²
This reduces to;
(1/2)m(v_f)² + (1/2)Ik(x_f)² = (1/2)k•(x_i)²
Making v_f the subject, we have;
v_f = [√(k/m)] * [√((x_i)² - (x_f)²)]
We know that ω = √(k/m)
Thus,
v_f = ω[√((x_i)² - (x_f)²)]
Plugging in the relevant values to obtain;
v_f = 17.8[√((0.068)² - (0.034)²)]
v_f = 17.8[0.059] = 1.05 m/s
Answer:
2.56 N*m
Explanation:
Torque is force times distance, with the force being perpendicular to the distance.
If the force and distance are not perpendicular, a projection of one of the two must be used instead.
We will use the projection of the length of the shaft upon a perpendicular of the force applied. This will have a magnitude of:
Lh = L * cos(a) (I name it Lh because it is a horizontal projection in this case)
Lh = 0.2 * cos(π/5) = 0.16 m
Then:
T = f * d
T = 16 * 0.16 = 2.56 N*m
5.8x10^3
3.02x10^8
4.5x10^5
8.6x10^10
Answer:
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
Explanation:
Given :
Frequencies of the sinusoids,
, and
Sampling rate 
The positive frequencies at the output of the sampling system are :
When n = 0,
when n = 1,
When n = 2,
Therefore, the first six positive frequencies present in the replicated spectrum are :
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
Answer:
B
Explanation:
Newton's second law:
F = ma
The acceleration is Δv/Δt, so:
F = m Δv/Δt
Given m = 1300 kg, Δv = 15 m/s, and Δt = 2 s:
F = (1300 kg) (15 m/s) / (2 s)
F = 9750 N
Answer B.
