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:
1.5024
Explanation:
Draw a diagram. Put the two cells in series. Now draw 3 resistors. Two of them equal 0.26 ohms each. The third one is the lightbulb which is 12 ohms.
R = 0.26 + 0.26 + 12 = 12.52
The bulb has a voltage of 2.88 volts across it. You can get the current from that.
i = E / R
i = 2.88 / 12 =
i = 0.24 amps.
Now you can get the voltage drop across the two cells.
E = ?
R = 0.26
i = 0.24 amps
E = 0.26 * 0.24
E = 0. 0624
Finally divide the 2.88 by 2 to get 1.44
Each cell has an emf of 1.44 + 0.0624 = 1.5024
Answer:
it would be 39.2 m/s
Explanation:
After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on.
Answer:
12.50 m/s
Explanation:
Vi = 9.49 m/s
a = 0.988 m/s²
t = 3.05 s
Vf = ?
Vf = Vi + at
Vf = 9.49 + (0.988)(3.05)
Vf = 12.50 m/s