Explanation:
The given data is as follows.
q = 6.0 nC = 
inner radius (r) = 1.0 cm = 0.01 m (as 1 cm = 100 m)
So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.
Formula to calculate the charge density is as follows.
.......... (1)
Since, area of the sphere is as follows.
A = 
........... (2)
Hence, substituting equation (2) in equation (1) as follows.
= 
= 
or, = 4.77 
Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 .
Answer:
The induced current and the power dissipated through the resistor are 0.5 mA and 
.
Explanation:
Given that,
Distance = 1.0 m
Resistance = 3.0 Ω
Speed = 35 m/s
Angle = 53°
Magnetic field 
(a). We need to calculate the induced emf
Using formula of emf
Where, B = magnetic field
l = length
v = velocity
Put the value into the formula
We need to calculate the induced current
Put the value into the formula
(b). We need to calculate the power dissipated through the resistor
Using formula of power
Put the value into the formula
Hence, The induced current and the power dissipated through the resistor are 0.5 mA and .
Answer:
Explanation:
Average speed = Total distance / Total time.
100 km/hr
r = 100 km / hr
t = 6 hours
d = 6 * 100 = 600 km
120 km / hr
r = 120 km / hr
t = 5 hour
d = 120 * 5
d = 600 km
Total distance = 600 + 600 = 1200 km
Total time = 5 hour + 6 hours = 11 hours.
Average speed = 1200 km / 11 hours = 109.1
Given:
mass: 100 kg
height: 500 m
1 kJ = 1000 J
gravity = 9.8 m/s²
velocity before impact: v = √2gh ; v = √2 * 9.8 m/s² * 500 m ; v = 98.99494 m/s
KE = 1/2 m v²
KE = 1/2 * 100 kg * (98.99494 m/s)²
KE = 490,000 J
Pls. see attachment.
Answer: 0.01 m
Explanation: The formulae for capillarity rise or fall is given below as
h = (2T×cosθ)/rpg
Where θ = angle mercury made with glass = 50°
T = surface tension = 0.51 N/m
g = acceleration due gravity = 9.8 m/s²
r = radius of tube = 0.5mm = 0.0005m
p = density of mercury.
h = height of rise or fall
From the question, specific gravity of density = 13.3
Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³
Hence density of mercury = 13.3×1000 = 13,300 kg/m³.
By substituting parameters, we have that
h = 2×0.51×cos 50/0.0005×9.8×13,300
h = 0.6556/65.17
h = 0.01 m
