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makkiz [27]
3 years ago
5

A 52 kg woman cheats on her diet and eats a 540 kcal jelly donut for breakfast. (a) How many joules of energy are the equivalent

of one jelly doughnut? (b) How many stairs must the woman climb to perform an amount of mechanical work equivalent to the food energy in one jelly doughnut?
Physics
1 answer:
Nadusha1986 [10]3 years ago
4 0

Answer:

a)801KJoules

b)The woman must climb stairs for an equivalent of 1.57km

Number of stairs=7850 stairs

Explanation:

a)We use the conversion between Joules and Kcal:

4184Joules=1Kcal

540Kcal*(1484Joules/1Kcal)=801KJoules

b)The physical Work must be equal to the food energy

W_{physical}=E_{food}\\ mgh=E_{food}\\h=E_{food}/(mg)=801KJoules/(52Kg*9.81m/s^{2})=1.57Km\\

The woman must climb stairs for an equivalent of 1.57Km

For example if an stair measure 20cm=0.2m

Number of stairs=1.57km/0.2m=7850 stairs

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Answer:

"2Ω" is the net resistance in the circuit.

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On substituting the values, we get

⇒  \frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}

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Electromagnetic waves are ........... by shiny surfaces.
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3 years ago
A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W
Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s

Since, the motion is downward, final velocity must be negative. So,

v_d=-4.95\ m/s

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Given:

Displacement of ball (S) = 0.665 m

Initial velocity (v_d) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s

Since, the motion is upward, final velocity must be positive. So,

v_{up}=3.31\ m/s

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

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