The buoyant force or upward buoyancy force
Answer: The velocity with which the sand throw is 24.2 m/s.
Explanation:
Explanation:
acceleration due to gravity, a = 3.9 m/s2
height, h = 75 m
final velocity, v = 0
Let the initial velocity at the time of throw is u.
Use third equation of motion
The velocity with which the sand throw is 24.2 m/s.
Answer:
x = 0.6034 m
Explanation:
Given
L = 5 m
Wplank = 225 N
Wman = 522 N
d = 1.1 m
x = ?
We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.
∑τ = 0 ⇒ τ₁ + τ₂ = 0
Torque come from the weight of the plank = τ₁
Torque come from the weight of the man = τ₂
⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)
⇒ τ₂ = Wman*x = 522 N*x (clockwise)
then
τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0
⇒ x = 0.6034 m
It is A or D but I believe A
The Pacific is loosely shaped like a triangle, opening widely to the south but barely at all to the north, while the Atlantic is shaped like an hourglass with the choke point located very loosely at the equator (somewhat south of it in the west).
