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2 years ago

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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

1 answer:

almond37 [142]2 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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Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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