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Art [367]
3 years ago
13

Qqqqqqqqqqqqqqqqqqqqqqqqqqqq

Physics
1 answer:
tester [92]3 years ago
3 0
Great question the answer is -25x.
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A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

7 0
3 years ago
The eiffel tower is a steel structure whose height increases by 19.5 cm when the temperature changes from −8 to +42 °c. what is
Readme [11.4K]
 The coefficient of expansion is 13 * 10^-6 m per meter length.per oK 
The temperature difference = 42 - - 8 = 50 oC 
delta T = (42 + 273) - (-8 + 273) = 50 oK 
delta L = L * 13* 10^6 m/oK 
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m 
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m 
6 0
3 years ago
Vector a cross b whole square<br><br>​
skad [1K]

Answer:

Kriss go Brrrrrrrrrrr

Explanation:

4 0
3 years ago
15 PTS!!!!
Anon25 [30]
I thinks its period because they are on the last column to the right
6 0
3 years ago
Read 2 more answers
The force applied when using a simple machine ??
beks73 [17]
Answer : I hope this helps !

The Effort Force is the force applied to a machine. Work input is the work done on a machine. The work input of a machine is equal to the effort force times the distance over which the effort force is exerted.
8 0
3 years ago
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