A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
1 answer:
You might be interested in
Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ =
So,
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Answer:
Explanation:
Given:
- mass of water,
- initial temperature of water,
- initial temperature of pan,
- mass of pan,
- mass of water evapourated,
- specific heat of water,
- specific heat of aluminium pan,
- latent heat of vapourization,
<u>Using the equation of heat:</u>
<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>