The solution for this problem would be:(10 - 500x) / (5 - x)
so start by doing:
x(5*50*2) - xV + 5V = 0.02(5*50*2)
500x - xV + 5V = 10
V(5 - x) = 10 - 500x
V = (10 - 500x) / (5 - x)
(V stands for the volume, but leaves us with the expression for x)
Answer:
A 93%
Explanation:
= Pressure will be equal at inlet and outlet
= Density of water = 1000 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
= Velocity at inlet = 1.2 m/s
= Velocity at outlet
= Radius of inlet = 
= Radius of outlet
From Bernoulli's relation

From continuity equation

The fraction would be

The fraction is 93.0304%
Answer:
F = 10.788 N
Explanation:
Given that,
Charge 1, 
Charge 2, 
Distance between charges, d = 0.1 m
We know that there is a force between charges. It is called electrostatic force. It is given by :

So, the force applied between charges is 10.788 N.
Momentum before the collision
x-direction:
p = m₁v₁ = 1.5 * 4.5 = 6.75
x-direction:
p = 0
momentum after the collision is conserved:
x-direction:
p = 6.75 = m₁v₁ + m₂v₂ = 1.5 * 2. 1* cos -30° + 3.2 * v₂*cos θ
y-direction:
p = 0 = m₁v₁ + m₂v₂ = 1.5 * 2.1 * sin -30° + 3.2 * v₂ * sin θ
Solve the two equations for v₂ and θ.
Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.
g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.
First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s
Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s
The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.
Answer:
First dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 8.42 m/s downward
Second dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 7.99 m/s downward
The landing velocities differ by 0.43 m/s.