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DerKrebs [107]
3 years ago
13

(a) The average mpg rating for an older model of a car is 24.8 miles to the gallon and it takes 6.60 x 10^5 J of work done on th

e car to drive one mile. Given that a gallon of gasoline gives 1.19 x 10^8 J of internal energy to the car, what is the amount of heat released to the atmosphere for each mile you drive the car?
Physics
1 answer:
Margarita [4]3 years ago
6 0

Answer:

the gasoline releases Q=-4138387 J/mile to the atmosphere

Explanation:

from the fist law of thermodynamics

ΔU = Q - W

where ΔU= change in internal energy , Q= heat received , W= work done by the system

then

Q = ΔU + W

knowing that

ΔU = internal energy per gallon * number of gallons consumed in 24.8 miles

= (-1.19 x 10^8 J/ gallon) * 1 gallon /24.8 miles = (-4798387 J/mile)

W = work done by mile= 6.60 x 10^5 J/mile

therefore the heat released per mile

Q = ΔU + W = -4798387 J/mile + 6.60 x 10^5 J/mile = -4138387 J

since Q is negative , energy is released into the atmosphere

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skad [1K]

Answer:

Length L=0.0125 m

Explanation:

Given data

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Temperature T=20°C

Resistivity of tungsten at 20°C p=5.6×10⁻⁸Ω.m²

Radius of filament r=0.09/2 =0.045 mm =0.045×10⁻³m

To find

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Solution

The Cross section of area of filament is given as:

A=\pi r^{2}\\ A=\pi (0.045*10^{-3})^{2}\\A=6.3585*10^{-9}m^{2}

The Resistance of a material is given as:

R=p\frac{L}{A} \\L=\frac{AR}{p}\\ L=\frac{(6.3585*10^{-9})(0.11)}{5.6*10^{-8}}\\L=0.125m

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3 years ago
When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s . By how much has thermal energy increased duri
Advocard [28]

Complete question:

A fireman of mass 80 kg slides down a pole. When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s. By how much has thermal energy increased during his slide?

Answer:

The thermal energy increased by 3,099.2 J

Explanation:

Given;

mass of the fireman, m = 80 kg

initial position of the fireman, hi = 4.2 m

final speed, v = 2.2 m/s

The change in the thermal energy is calculated as;

ΔE +  (K.Ef - K.Ei) + (Uf - Ui) = 0

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ΔE is the change in the thermal energy

K.Ef is the final kinetic energy

K.Ei is the initial kinetic energy

Uf is the final potential energy

Ui is the initial potential energy

\Delta  E_{th} +  (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i)=0\\\\initial \ velocity, \ v_i = 0\\final \ height , \ h_f = o\\\\\Delta  E_{th} + (\frac{1}{2} mv_f^2) + ( - mgh_i)=0\\\\\Delta  E_{th} +  \frac{1}{2} mv_f^2 - mgh_i = 0\\\\\Delta  E_{th} =mgh_i  - \frac{1}{2} mv_f^2\\\\\Delta  E_{th} =  80 \times 9.8 \times 4.2  \ \ - \ \ \frac{1}{2} \times 80 \times (2.2)^2  \\\\\Delta  E_{th} = 3292.8 \ J \ - \ 193.6 \ J\\\\\Delta  E_{th} = 3,099.2 \ J

3 0
3 years ago
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