Answer:
u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)
Explanation:
Given:
- Electric Field strength near earth's surface E = 145 V / m
- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg
Find:
- How much energy is stored per cubic meter in this field?
Solution:
- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:
u_e = 0.5*e_o * E^2
- Plug in the values given:
u_e = 0.5*8.854 *10^-12 *145^2
u_e = 9.30777 * 10^-8 J/m^3
Answer:
To help wheels move in a circle
Explanation:
Answer:
the heat transfer from the pipe will decrease when the insulation is taken off for r₂< 
where;
r₂ = outer radius
= critical radius
Explanation:
Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .

The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than critical radius (r₂<
) 0, and reaches a maximum when r₂ =
, and starts to decrease for r₂<
. Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂<
.
Answer:
Temperature
Explanation:
In an ideal gas the specific enthalpy is exclusively a function of Temperature only this can be also written as h = h(T)
A gas is said be ideal gas if obeys PV= nRT law
And in a ideal gas both internal energy and specific enthalpy are a function of Temperature only. Therefore the constant volume and constant pressure specific heats Cv and Cp are also function of temperature only.
Answer:
a.)
US Sieve no. % finer (C₅ )
4 100
10 95.61
20 82.98
40 61.50
60 42.08
100 20.19
200 6.3
Pan 0
b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4
c.) Cu = 3.33
d.) Cc = 1
Explanation:
As given ,
US Sieve no. Mass of soil retained (C₂ )
4 0
10 18.5
20 53.2
40 90.5
60 81.8
100 92.2
200 58.5
Pan 26.5
Now,
Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g
⇒ w = 421.2 g
As we know that ,
% Retained = C₃ = C₂×
∴ we get
US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)
4 0 0
10 4.39 4.39
20 12.63 17.02
40 21.48 38.50
60 19.42 57.92
100 21.89 79.81
200 13.89 93.70
Pan 6.30 100
Now,
% finer = C₅ = 100 - C₄
∴ we get
US Sieve no. Cummulative % retained (C₄) % finer (C₅ )
4 0 100
10 4.39 95.61
20 17.02 82.98
40 38.50 61.50
60 57.92 42.08
100 79.81 20.19
200 93.70 6.3
Pan 100 0
The grain-size distribution is :
b.)
From the diagram , we can see that
D10 = 0.12
D30 = 0.22
D60 = 0.12
c.)
Uniformity Coefficient = Cu = 
⇒ Cu = 
d.)
Coefficient of Graduation = Cc = 
⇒ Cc =
= 1