Answer:
Explanation:
We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .
∫ E ds = q / ε
Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so
∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .
This also represents total flux coming out of the charge q on all sides .
This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .
If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .
It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors
1 ) charge q and
2 ) the permittivity of medium ε around .
Answer:
126.99115 g
Explanation:
50 g at 90 cm
Stick balances at 61.3 cm
x = Distance of the third 0.6 kg mass
Meter stick hanging at 50 cm
Torque about the support point is given by (torque is conserved)
The mass of the meter stick is 126.99115 g
Answer:
the value of the final pressure is 0.168 atm
Explanation:
Given the data in the question;
Let p₁ be initial pressure, v₁ be initial volume.
After expansion, p₂ is final pressure and v₂ is final volume.
So using the following equations;
p₁v₁ = nRT
p₂v₂ = nRT
hence, p₁v₁ = p₂v₂
we find p₂
p₂ = p₁v₁ / v₂
given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,
final volume v₂ = 0.365 m³
we substitute
p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³
p₂ = 0.06125 atm-m³ / 0.365 m³
p₂ = 0.168 atm
Therefore, the value of the final pressure is 0.168 atm
