This occurs when a person comes into contact with hazard
1 answer:
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Answer:
find the diagram in the attachment.
Explanation:
Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.
considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:
(vf)^2 = (vi)^2 + 2×g×Δy
vf = 0 m/s, at the highest point in the upward motion, then:
0 = (vi)^2 + 2×g×Δy
-(vi)^2 = 2×g×Δy
Δy = [-(vi)^2]/2×g
Δy = [-(-12)^2]/(2×9.8)
Δy = - 7.35 m
then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown
Answer:
The buoyant force is 3778.8 N in upward.
Explanation:
Given that,
Mass of balloon = 222 Kg
Volume = 328 m³
Density of air = 1.20 kg/m³
Density of helium = 0.179 kg/m³
We need to calculate the buoyant force acting
Using formula of buoyant force
Where, = density of air
V = Volume of balloon
g = acceleration due to gravity
Put the value into the formula
This buoyant force is in upward direction.
Hence, The buoyant force is 3778.8 N in upward.