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2 answers:
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Consult the attached free body diagram.
By Newton's second law, the net force on the crate acting parallel to the surface is
∑ F[para] = (370 N) cos(-20°) - f = 0
(this is the x-component of the resultant force)
where
• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force
• f = magnitude of kinetic friction
The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.
Solve for f :
f = (370 N) cos(-20°) ≈ 347.686 N
The net force acting perpendicular to the surface is
∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0
(this is the y-component of the resultant force)
where
• n = magnitude of normal force
• 1480 N = weight of the crate
• (370 N) sin(-20°) = magnitude of the vertical component of push
The crate doesn't move up or down, so it's also in equilibrium in this direction.
Solve for n :
n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N
Then the coefficient of kinetic friction is µ such that
f = µn ⇒ µ = f/n ≈ 0.216
Answer:
32 cm³
Explanation:
The given gas data are;
The relative density of oxygen = 16
The relative density of carbon dioxide = 12
The time it takes 25 cm³ of carbon dioxide to effuse out = 75 seconds'
The duration of effusion of the oxygen = 96 seconds
The rate of effusion of carbon dioxide, R1 = 25 cm³/(75 sec) = (1/3) cm³/sec
According to Graham's law of diffusion and effusion of a gas, we have;
Therefore, we have;
The volume of effusion = The rate of effusion × Time
The volume of the oxygen that will effuse in 96 seconds is given as follows;
The rate of effusion of a gas × Time
V = The rate of effusion of oxygen × Time = (1/3) cm³/sec × 96 sec = 32 cm³
The volume of oxygen that will effuse in 96 seconds, V = 32 cm³.