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Sophie [7]
3 years ago
11

Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%

imbalance in the amount of positive and negative charge, one student being positive and the other negative. Estimate the electrostatic force of attraction between them by replacing each student with a sphere of water having the same mass as the student.
Physics
1 answer:
djyliett [7]3 years ago
4 0

Answer:

6.8370869499\times 10^{20}\ N

Explanation:

N_A = Avogadro's number = 6.022\times 10^{23}

e = Charge of electron = 1.6\times 10^{-19}\ C

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

q=imbalance\times n\times e

Total number of protons and electrons in each sphere

n=\dfrac{mN_AZe}{M}

q=imbalance\times \dfrac{mN_AZe}{M}

q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C

q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C

Electrical force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N

The electrostatic force of attraction between them is 6.8370869499\times 10^{20}\ N

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Answer:

You will hear the note E₆

Explanation:

We know that:

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The Doppler effect says that:

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v = velocity of the sound wave

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vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.

Replacing the values that we know in the equation we have:

f' = \frac{340 m/s + 88m/s}{340 m/s} *1,046 Hz = 1,316.73 Hz

This frequency is close to the note E₆ (1,318.5 Hz)

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2 years ago
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8 0
3 years ago
You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
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a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.

from above statement we got

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since the ball is thrown, so its vertical velocity would be zero

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taking g = 9.8m/s^2

now, using the equation of motion

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t = 4 sec

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range =  20 m

c)  vertical components of the stone’s velocity just before it hits the ground = v sin θ =

horizontal   components of the stone’s velocity  just before it hits the ground = v cos θ

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