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konstantin123 [22]
3 years ago
9

Find the total electric charge of 2.5 kg of electrons. Express your answer using two significant figures.

Physics
1 answer:
zimovet [89]3 years ago
4 0

Answer : The total electric charge of electrons is, -4.4\times 10^{11}C

Explanation:

Answer : The number of electrons transferred are, 4.68\times 10^{20}

Explanation :

First we have to calculate the number of electrons.

Number of electrons = \frac{\text{Total mass of electrons}}{\text{Mass of one electron}}

Mass of 1 electron = 9.1\times 10^{-31}kg

Total mass of electron = 2.5 kg

Number of electrons = \frac{2.5kg}{9.1\times 10^{-31}kg}

Number of electrons = 2.75\times 10^{30}

Now we have to calculate the total electric charge of electrons.

Formula used :

Q=ne\\\\n=\frac{Q}{e}

where,

n = number of electrons transferred = 2.75\times 10^{30}

Q = charge on electrons = ?

e = charge on 1 electron = -1.602\times 10^{-19}C

Now put all the given values in the above formula, we get:

2.75\times 10^{30}=\frac{Q}{-1.602\times 10^{-19}C}

Q=-4.4\times 10^{11}C

Thus, the total electric charge of electrons is, -4.4\times 10^{11}C

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tresset_1 [31]

Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, L=50\ mH=0.05\ H

Voltage source, V = 15 volt

Frequency, f = 400 Hz

The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :

X_L=2\pi fL

X_L=2\pi \times 400\ Hz\times 0.05\ H

X_L=125.66\ \Omega

X_L=125.7\ \Omega

So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

8 0
3 years ago
Insulators will:
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Answer:

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Explanation:

An electric current usually consists of electrons moving through a wire.

An insulator prevents the flow of an electric current, so it inhibits the flow of electrons.

7 0
4 years ago
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A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o
Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m)   ⇒   h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv²   ⇒   d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0
3 years ago
WHY does an acceleration cause velocity to increase
harkovskaia [24]

Answer:

When the velocity of an object changes it is said to be accelerating. Acceleration is the rate of change of velocity with time. ... Acceleration occurs anytime an object's speed increases or decreases, or it changes direction.

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A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i
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Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

\boxed{\mathfrak{range =  \frac{  u  {}^{2}   \sin 2\theta }{g} }}

u is the velocity during its take off and \theta is the angle at which its thrown

Given that

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  • \theta = 40°

calculating range using the above formula

= \frac{ {8}^{2} \sin2(40)  }{10}

=  \frac{64 \times  \sin(80) }{10}

value of sin 80 = 0. 985

=  \frac{64 \times 0.985}{10}

=  \frac{63.027}{10}

= 6.3027

Hence,

\mathfrak { \blue{the \: teammate \: is \:  \red{\underline{6.3027 \: meters} }\: away } }

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