The acceleration due to gravity on the surface of the moon is 1.62m/s^2. The moon's radius is Rm+1738km. A) What is the weight i
1 answer:
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Answer:
The capacity of the highway section = 400 vehicle/hr
The capacity of the highway section = 40 km/hr
The density when the highway is at one-quarter of the capacity = 13.39 vehicle/km
Explanation:
See attached pictures.
Answer:
- minimum separation distance between the two legs of the sting L = L 1 + L 2 therefore L = 9.48 + 4.68 = 14.16 m
- L = 1.14 m
Explanation:
D ( diameter ) = 125 m
convection coefficient of h = 700 W/m^2
Calculate THE CROSS SECTIONAL AREA
Ac = =
= 0.79 * 15625 = 12343.75 m^2
perimeter
p = = 3.14 * 125 = 392.5 m
at 300k temperature the thermal conductivity of copper and constantan from the thermodynamic property table are :
Kcu = 401 w/m.k
Kconstantan = 23 W/m.k
To calculate the length of copper wire of the thermocouple junction
L 1 = 4.6 () ^ 1/2 = 4.6
L 1 = 4.6 ( 4949843.75 / 274750 )^1/2
L 1 = 9.48 m
calculate length of constantan wire
L 2 = 4.6
= 4.6 ( (23 * 12343.75) / ( 700 * 392.5) ) ^1/2
L 2 = 4.6 ( 283906.25 / 274750 ) ^ 1/2
L 2 = 4.68 m
I) therefore the minimum separation distance between the two legs of the sting L = L 1 + L 2
L = 9.48 + 4.68 = 14.16 m
ii) Evaluating the thermal conductivity of copper and constantan
Kc ( thermal conductivity of chromel) = 19 w/m.k
Ka ( thermal conductivity of alumel ) = 29 W/m.k
distance between the legs L = L 1 + L 2
THEREFORE
L = 4.6 ( (Kcn * Ac ) / ( hp ) )^1/2 + 4.6 ( (Kac * Ac)/(hp) )^1/2
L = 4.6
L = 4.6 ( 12343.75 /( 700 * 392.5) )^1/2 * [ 19^1/2 + 29^1/2 ]
L = 4.6 ( 12343.75 / 274750 ) ^1/2 * 5.39
L = 1.14 m
Answer:
The Current will decrease by a factor of 2
Explanation:
Given the conditions, it should be noted that the current in the circuit is determined by the LOAD. In other words, the amount of current generator will be producing depends upon the load connected to it.
Now, as the question says, the load is reduced to half its original value, we can write:
Since, P2 = P1/2,
Dividing equations (1) and (2), we get,
P1 / (P1/2) = I1/ I2
Hence, it is proved that the current in the transmission line will decrease by a factor of 2 when load is reduced to half.