Answer:
see explaination
Explanation:
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- The conversion of 47,000 Ohms to kilo-ohms is equal to 47 kilo-ohms.
- The conversion of 47,000 Ohms to mega-ohms is equal to 0.047 kilo-ohms.
<h3>What is
resistance?</h3>
Resistance can be defined as an opposition to the flow of current in an electric circuit. Also, the standard unit of measurement of the resistance of an electric component is Ohms, which can be converted to kilo-ohms or mega-ohms.
For Ohms to kilo-ohms, we have:
1 Ohms = 0.001 kilo-ohms
47,000 Ohms = X kilo-ohms
Cross-multiplying, we have:
X = 0.001 × 47000
X = 47 kilo-ohms.
For Ohms to mega-ohms, we have:
1,000,000 ohms = 1 mega-ohms
47,000 Ohms = X mega-ohms
Cross-multiplying, we have:
X1,000,000 = 47,000
X = 47,000/1,000,000
X = 0.047 kilo-ohms.
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Answer:
0.71 lbf
Explanation:
Use ideal gas law:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.
P = 32 lbf/in² + 14.7 lbf/in²
P = 46.7 lbf/in²
Absolute temperature is in Kelvin or Rankine:
T = 75 + 459.67 R
T = 534.67 R
Given V = 3.0 ft³, and R = 10.731 ft³ psi / R / lb-mol:
PV = nRT
(46.7 lbf/in²) (3.0 ft³) = n (10.731 ft³ psi / R / lb-mol) (534.67 R)
n = 0.02442 lb-mol
The molar mass of air is 29 lbm/lb-mol, so the mass is:
m = (0.02442 lb-mol) (29 lbm/lb-mol)
m = 0.708 lbm
The weight of 1 lbm is lbf.
W = 0.708 lbf
Rounded to two significant figures, the weight of the air is 0.71 lbf.
Answer:
Rate of heat transfer is 0.56592 kg/hour
Explanation:
Q = kA(T2 - T1)/t
Q is rate of heat transfer in Watts or Joules per second
k is thermal conductivity of the styrofoam = 0.035 W/(mK)
A is area of the cubical picnic chest = 6L^2 = 6(0.5)^2 = 6×0.25 = 1.5 m^2
T1 is initial temperature of ice = 0 °C = 0+273 = 273 K
T2 is temperature of the styrofoam = 25 °C = 25+273 = 298 K
t is thickness of styrofoam = 0.025 m
Q = 0.035×1.5(298-273)/0.025 = 1.3125/0.025 = 52.5 W = 52.5 J/s
Mass flow rate = rate of heat transfer ÷ latent heat of melting of ice = 52.5 J/s ÷ 3.34×10^ 5 J/kg = 1.572×10^-4 kg/s = 1.572×10^-4 kg/s × 3600 s/1 hr = 0.56592 kg/hr
Answer:
Super critical
1.2 m
Explanation:
Q = Flow rate = 
w = Width = 3 m
d = Depth = 90 cm = 0.9 m
A = Area = wd
v = Velocity
g = Acceleration due to gravity = 
Froude number is given by
Since 
the flow is super critical.
Flow is critical when 
Depth is given by
The depth of the channel will be 1.2 m for critical flow.
