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statuscvo [17]
2 years ago
4

A 15.53 kg loose bag of soil falls 5.50 m at a construction site. If all the potential energy of the fall is retained by the soi

l in the bag, how much will its temperature increase? (csoil = 0.200 kcal/kg oC)
Physics
1 answer:
nignag [31]2 years ago
5 0

Answer:

The rise in temperature is 0.064 ⁰C

Explanation:

Given;

mass of  loose bag of soil, m = 15.53 kg

height of fall, h = 5.50 m

Potential energy of the fall, U = mgh

                                              U = 15.53 x 9.8 x 5.5

                                              U = 837.067 J = 0.200 kCal

If all the potential energy of the fall is retained by the soil in the bag, then the potential energy of fall is equal to thermal energy of the soil.

U = Q = mcΔθ

where;

Q is quantity of heat  retained by the soil in the bag

m is mass of the loose bag of soil

c is specific heat capacity of the soil =  0.200 kcal/kg ⁰C

Δθ is increase in temperature

Δθ  =  U / mc

Δθ  =  0.2 / (15.53 x 0.2)

Δθ  = 0.064 ⁰C

Therefore, the rise in temperature is 0.064 ⁰C

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Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be a
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The movable piston

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8 0
2 years ago
Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The
Eddi Din [679]

Answer:

(a) Steel rod: 1.1 * 10^{-4}

    Copper rod: 1.88 * 10^{-4}

(b) Steel rod: 8.3 * 10^{-5} m

Copper rod: 1.41 * 10^{-4} m

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = \frac{1}{Y} * \frac{F}{A}

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y =  200 000 000 000 N/m^{2}

F = 4000N

A = \pi r^{2}      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = \pi * (0.0075)^{2}

=> A = 0.000177 m^{2}

∴ Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}

For the copper rod:

Y =  120 000 000 000 N/m²

F = 4000N

A = \pi r^{2}      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = \pi * (0.0075)^{2}

=> A = 0.000177 m^{2}

Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain =  = 1.88 * 10^{-4}

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = 1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m

For the copper rod:

Elongation = 1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m

6 0
2 years ago
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