You throw a 20-n rock vertically into the air from ground level. you observe that when it is a height 16.0 m above the ground, i
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:
A) B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West
Explanation:
A) For this exercise we must use the expression of Faraday's law for a moving body
fem =
fem = - d (B l y) / dt = - B lv
B =
we calculate
B = - 7.9 10⁻⁴ /(0.73 20)
B = 5.4 10⁻⁵ T
B) to determine which side of the bar is positive, we must use the right hand rule
the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.
Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West