A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by
Paul has magnitude 43.0 N and direction 61.0∘ south of west.
How much work does Paul's force do during a displacement of the crate that is 12.0 m in the direction 22.0∘ east of north?
How much work does Paul's force do during a displacement of the crate that is 12.0 m in the direction 22.0∘ east of north?
1 answer:
Answer:
W = -510.98J
Explanation:
Force = 43N, 61° SW
Displacement = 12m, 22° NE
Work done is given as:
W = F*d*cosA
where A = angle between force and displacement.
Angle between force and displacement, A = 61 + 90 + 22 = 172°
W = 43 * 12 * cos172
W = -510.98J
The negative sign shows that the work done is in the opposite direction of the force applied to it.
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Question
Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second.
Answer:
Explanation:
The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,
Here, l is the length of the baton.
Substitute 0.120 kg for m, 3 rads/s for
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
0.69444 m, 0.08152 m, 0.32407 m, 0.03804 m
Explanation:
v = Velocity of sound
f = Frequency
Length of vocal tract is given by
At f = 270 Hz v = 750 m/s
At f = 2300 Hz v = 750 m/s
At f = 270 Hz v = 350 m/s
At f = 2300 Hz v = 350 m/s