Answer:
Explanation:
It is given that,
Charge at origin,
Let second charge is and it is placed at x = 0.1 m
Resulting force on the second charge,
The electric force acting on the charged particles is given by the following formula as :
So, the second charge is . Hence, this is the required solution.
Answer:
1767Kw
Explanation:
Velocity of wind = 10 m/s
diameter of the blades= 60m
ρ= air density = 1.25 kg/m3
Acceleration due to gravity= 9.81 m/s^2
Mechanical energy of the wind can be calculated using the expression below
Energy= (e*m)
= ρ V A e............eqn(1)
Where A= area
ρ= air density
e= wind energy per unit mass of air
e= (v^2)/2..........eqn(2)
If we substitute the values into eqn (2) we have
e= [(10)^2]/2
=50J/Kg
But Area=A= (πd^2)/4
Area= ( π× 60^2)/4
Area=2827.8m^2
If we input substitute the values into eqn (1) we have
Energy= 1.25 ×10 × 50×2827.8
=1767145.7W
We can convert to kilo watt
=1767145.7W/ 1000
= 1767Kw
Hence, the mechanical energy of air per unit mass and the power generation potential of a wind turbine is 1767Kw
I will run 2 miles three times a week for 4 weeks.
Answer:
3333.33 kg/m³
Explanation:
Density: This can be defined as the ratio of the mass of a body to its volume.
The unit of density is kg/m³.
From Archimedes principle,
R.d = W/U = D/D'
Where R.d = relative density, W = weight of the object in air, u = upthrust in water, D = Density of the object, D' = Density of water.
W/U = D/D'
making D the subject of the equation
D = D'(W/U).......................... Equation 1
Given: W = 5.0 N, U = 5.0 -3.5 = 1.5 N, D' = 1000 kg/m³
Note: U = lost in weight = weight in air - weight in water
Substitute into equation 1
D = 1000(5/1.5)
D = 3333.33 kg/m³
Thus the density of the object = 3333.33 kg/m³