Two teams are playing tug-of-war. Team A, on the left, is pulling on the rope with an effort of 5000 N. If the rope is moving at
1 answer:
You might be interested in
Answer:
The force required to hold the contraction in place is 665.91 N ↑
Explanation:
Given;
specific gravity of oil, γ = 0.75
Volumetric flow rate, V 3.2 Ft³/s = 0.0906 m³/s
where;
is the density of oil
is the density of water = 1000 kg/m³
∴density of oil () = γ × density of water(
)
= 0.75 × 1000 kg/m³
= 750kg/m³
Buoyant Force = ρVg
= 750 × 0.0906 × 9.8
= 665.91 N ↑
This force acts upward or opposite gravitational force.
Therefore, the force required to hold the contraction in place is 665.91 N ↑
Answer:
Explanation:
Be M1 and M2 the two massive objects and m the third one
The gravitational force over the third object at the beginning is zero (assuming that the position of the third object is x=0)
when the third object is displaced we have that the new position is x=Δx. Hence
I attached a scheme of the system
I hope this is useful for you
Regards
