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Levart [38]
3 years ago
7

An object is given a very small amount of charge. Which of the following could

Physics
1 answer:
spayn [35]3 years ago
3 0

5.4*10^-19 C

Explanation:

For the purposes of this question, charges essentially come in packages that are the size of an electron (or proton since they have the same magnitude of charge). The charge on an electron is -1.6*10^-19

Therefore, any object should have a charge that is a multiple of the charge of an electron - It would not make sense to have a charge equivalent to 1.5 electrons since you can't exactly split the electron in half. So the charge of any integer number of electrons can be transferred to another object.

Charge = q(electron)*n(#electrons)

Since 5.4/1.6 = 3.375, we know that it can not be the right answer because the answer is not an integer.

If you divide every other option listed by the charge of an electron, you will get an integer number.

(16*10^-19 C)/(1.6*10^-19C) = 10

(-6.4*10^-19 C)/(1.6*10^-19C) = -4

(4.8*10^-19 C)/(1.6*10^-19C) = 3

(5.4*10^-19 C)/(1.6*10^-19C) = 3.375

(3.2*10^-19C)/(1.6*10^-19C) = 2

etc.

I hope this helps!

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Answer:

2.72 km

Explanation:

(12.33 km)/ 1 hr * (1 hr)/ 60 min

0.2055 km/ min

distance=rate * time (assuming v is constant,

a=0)

(0.2055 km/ min)*(13.22 min)

2.72 km OR 2716.71 m

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A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5
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Answer:

V = 38.0 L

Explanation:

As we know that number of moles will remains conserved inside the balloon

so we will have

moles = \frac{PV}{RT}

here we have

\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}

now we have

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P_2 = 76 mm Hg

V_1 = 5.00 L

T_1 = 20^o C = 293 K

T_2 = -50^o C = 223 K

\frac{(760mm Hg)(5L)}{R(293)} = \frac{(76mm Hg)(V)}{R(223)}

V = 38.0 L

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Phosphorus bursts into flame when exposed to oxygen. What element might be used when storing pure phosphorus to keep it from rea
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<h3><u>Answer;</u></h3>

Argon

<h3><u>Explanation;</u></h3>
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Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

x=4t^3+3t^2-5t+2

Where,

x is in meters and t is in sec

We know that,

Velocity,

v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5

(a) i. t = 2 s

v=12(2)^2+6(2)-5=55\ m/s

At t = 4 s

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(b) Acceleration,

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