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harkovskaia [24]
3 years ago
6

A rigid tank internal energy of fluid 800kJ. Fluid loses 500kJ of heat and padle wheel does 100kJ of work. Find final internal e

nergy in tank.
Physics
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

 U₂ = 400 KJ      

Explanation:

Given that

Initial energy of the tank ,U₁= 800 KJ

Heat loses by fluid ,Q= - 500 KJ

Work done on the fluid ,W= - 100 KJ

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take final internal energy =U₂

We know that

Q= U₂ - U₁ + W

-500 = U₂ - 800 - 100

U₂ = -500 +900 KJ

U₂ = 400 KJ

Therefore the final internal energy = 400 KJ

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3 years ago
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height
rewona [7]

Answer:

W=17085KJ

Explanation:

From the question we are told that:

Height H=16m

Radius R=3

Height of water H_w=9m

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Generally the equation for Volume of water is mathematically given by

 dv=\pi*r^2dy

 dv=\frac{\piR^2}{H^2}(H-y)^2dy

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 dw=(pdv)g (H-y)

Substituting dv

 dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)

 dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy

Therefore

 W=\int dw

 W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy

 W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)

 W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0

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 W=17085KJ

 

'

'

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A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
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Answer:

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Explanation:

Resistivity is given by \rho=\frac {AR}{L} where A is cross-sectional area, R is resistance, L is the length and \rho is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

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3 years ago
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