12*8A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in
motion from the equilibrium position with a downward velocity of 2 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) . Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second
1 answer:
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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:
The magnitudes of the net magnetic fields at points A and B is 2.66 x T
Explanation:
Given information :
The current of each wires, I = 4.7 A
dH = 0.19 m
dV = 0.41 m
The magnetic of straight-current wire :
B= μI/2πr
where
B = magnetic field (T)
μ = 1.26 x
(N/
)
I = Current (A)
r = radius (m)
the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,
BH = μI/2πr
= (1.26 x )(4.7)/(2π)(0.19)
= 4.96 x T
BV = μI/2πr
= (1.26 x )(4.7)/(2π)(0.41)
= 2.3 x T
hence,
the net magnetic field = BH - BV
= 4.96 x - 2.3 x
= 2.66 x T
Answer:
shown in the attachment
Explanation:
The detailed step by step and necessary mathematical application is as shown in the attachment.
