Buoyancy is a force that always acts in an upward direction exerted by a fluid on a body placed in the fluid
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Answer:
5. Is greater than mg, always
Explanation:
If the cone has an inclination of angle β, the sum of forces will be:
x-axis (centripetal axis):
N*sin β = m*ax where ax is the centripetal acceleration
y-axis:
N*cos β - m*g = m*ay where ay is the vertical acceleration. If the block starts falling down, ay will be negative. If the block starts sliding up, ay will be positive. If the block does not move up nor down, ay=0.
Solving for N:

If ay is positive or zero, N will be greater than mg. If ay is negative, N will be less than mg.
If the block is sliding along a horizontal circular path (not up, nor down), ay = 0, so N will always be greater than mg.
The work done by a constant force in a rectilinear motion is given by:

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

Therefore, the total work done is 578.123 J and the answer is option E
Answer:
<u>We are given:</u>
initial velocity (u) = 20m/s
acceleration (a) = 4 m/s²
time (t) = 8 seconds
displacement (s) = s m
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<u>Solving for Displacement:</u>
From the seconds equation of motion:
s = ut + 1/2 * at²
replacing the variables
s = 20(8) + 1/2 * (4)*(8)*(8)
s = 160 + 128
s = 288 m
m1= mass 1 = 1.1 kg
Vi1 = initial velocity 1 = 2.7 m/s
m2= 2.4 kg
V2i = -1.9 m/s
We assume east as positive and west as negative.
Apply the formulas:
Vf1 = ?

Replacing:



Answer: 3.6 m/s west