The answer to your question is capacitor :)
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: SI Unit is meters per second(m/s)
Explanation:
The three units of speed are:
1. meters per second(m/s)
2. kilometers per hour(km/hr)
3. miles per hour(mph)
SI Unit is meters per second(m/s)
Freezing point of the water is known as 273 K
Hope this helps!
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 =
/ T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.