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:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:
5984.67N
Explanation:
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?
from continuity equation
v1A1=v2A2
equation of continuity
v1=4ft /s=1.21m/s
d1=14 inch=.35m
d2=14-2=0.304m
A1=pi*d^2/4
0.096m^2
a2=0.0706m^2
from continuity once again
1.21*0.096=v2(0.07)
v2=1.65
force on the pipe
(p1A1- p2A2) + m(v2 – v1)
from bernoulli
p1 + ρv1^2/2 = p2 + ρv2^2/2
difference in pressure or pressure drop
p1-p2=2psi
13.789N/m^2=rho(1.65^2-1.21^2)/2
rho=21.91kg/m^3
since the pipe is cylindrical
pressure is egh
13.789=21.91*9.81*h
length of the pipe is
0.064m
AH=volume of the pipe(area *h)
the mass =rho*A*H
0.064*0.07*21.91
m=0.098kg
(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)
force =5984.67N
If it is. DC, direct current reverse the polarity of power leads on the motor.
If it is a 3 phase ac alternating current, reverse any of the two of three leads.
Disconnect power before attempting.
Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
= 5 Mpa
= 1623°C
= 1896 K
= 0.05 
Also given 
Therefore, 
= 1
R = 0.27 kJ / kg-K
= 0.8 kJ / kg-K
Also given : 
Therefore, 
= 
= 0.1182 MPa
a). Work transfer, δW = 
![\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B5%5Ctimes%200.05-0.1182%5Ctimes%201%7D%7B1.25-1%7D%20%20%5Cright%20%5D%5Ctimes%2010%5E%7B6%7D)
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
= 
=![\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cfrac%7B%5Cgamma%20-n%7D%7B%5Cgamma%20-1%7D%20%5Cright%20%5D%5Ctimes%20%5Cdelta%20W)
=![\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cfrac%7B1.4%20-1.25%7D%7B1.4%20-1%7D%20%5Cright%20%5D%5Ctimes%20527.200)
= 197.7 kJ
