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Mrrafil [7]
3 years ago
6

A slice of bread contains about 100 kcal. If specific heat of a person were 1.00 kcal/kg·°C, by how many °C would the temperatur

e of a 70.0-kg person increase if all the energy in the bread were converted to heat?a. 2.25°Cb. 1.86°Cc. 1.43°Cd. 1.00°C
Physics
1 answer:
almond37 [142]3 years ago
5 0

Answer:

(c) 1.43°C

Explanation:

If the energy in the bread are converted to heat.

Then, The heat transferred from the bread to person = 100 kcal.

From specific heat capacity,

Q = cmΔT............................ equation 1

Where Q = quantity of heat, m = mass of the person, c = specific heat capacity of the person, Δ = increase in temperature.

Making ΔT the subject the equation 1,

ΔT = Q/cm........................ equation 2

Where Q = 100 kcal, c= 1.00 kcal/kg.°C, m = 70.0 kg

Substituting these values into equation 2,

ΔT = 100/(1×70)

ΔT = 100/70

ΔT = 1.428

ΔT ≈ 1.43°C

The increase in temperature of the body is = 1.43°C

The right option is (c) 1.43°C

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ivolga24 [154]

Answer:

The acceleration of the wallet is 3\hat{i}+6\hat{j}

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Given that,

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Using formula of acceleration

a=r\omega^2

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\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}

\dfrac{a}{r}=\dfrac{a'}{r'}

\dfrac{a'}{a}=\dfrac{r'}{r}

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a'=3\hat{i}+6\hat{j}

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3 years ago
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saveliy_v [14]

Answer:

0.00493 m/s

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The Root Mean Square speed is 0.00493 m/s

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Answer:

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substituting these in equation, we get:

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A(s)³ = m³

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Also,

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Answer:

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<h3><u>Answer;</u></h3>

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