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otez555 [7]
3 years ago
10

If 31.25

Physics
2 answers:
dlinn [17]3 years ago
5 0

Answer:

it will become charged by Q = 5C

Explanation:

When electron is removed from matter then the system will get positively charged

So we will have

Q = Ne

here N = number of electrons that is removed

N = 31.25 \times 10^{18}

also we know that charge on one electron is given as

e = 1.6 \times 10^{-19}

here we have

Q = Ne

Q = (31.25 \times 10^{18})(1.6 \times 10^{-19})

Q = 5 C

so it will become charged by Q = 5C

valentina_108 [34]3 years ago
4 0
<span>1 C = 6.24150965(16)×10^18 electrons

31.25 x 10^18 electrons / (6.24150965(16)×10^18 electrons / C) = 5.007 Coulombs

</span><span>I hope this helps. </span>
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Whats the question? ....
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A transverse wave on a string is described by the wave function
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The acceleration of the given wave is 2.5m/s^2

<h3>Wave property</h3>

The standard wave function is expressed according to the equation

y = Asin(2πft+2πx/λ)

where

λ is the wave length

Given the equation below

y = 0.120 sin((π/8)x + 4πt)

Compare both equation to have:

(π/8)x = 2πx/λ

8 = 2/λ

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λ = 1/4

λ = 0.25m

For the frequency

Compare both equation to have:

4πt = 2πft

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speed = 2(0.25)

speed = 0.5m/s

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acceleration = change in velocity/time

acceleration = 0.5/ 0.200

acceleration = 2.5m/s^2

Hence the acceleration of the given wave is2.5m/s^2

Learn more on wave function here: brainly.com/question/25699025

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7 0
2 years ago
A nonconducting ring with a radius of 11.5 cm is uniformly charged with a total positive charge of 10.0 µC. The ring rotates at
zhuklara [117]

Answer:

B=1.21*10^{-10}T

Explanation:

The magnitude of the magnetic field on the axis of the ring is given by:

B=\frac{\mu_0 IR^2}{2(r^2+R^2)^{\frac{3}{2}}}(1)

\mu_0 is the permeability of free space, I is the flowing current  through the ring, R is the ring's radius and r is the distance to the center of the ring.

The flowing current  through the ring is defined as the ring's charge divided into the time taken by the charge to complete one revolution, that is, the period T=\frac{2\pi}{\omega}. So, we have:

I=\frac{q}{T}\\I=\frac{q}{\frac{2\pi}{\omega}}\\\\I=\frac{\omega q}{2\pi}\\I=\frac{18\frac{rad}{s}(10*10^{-6}C)}{2\pi}\\I=2.87*10^{-5}A

Now, replacing in (1):

B=\frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(2.87*10^{-5}A)(0.115m)^2}{2((0.05m)^2+(0.115m)^2)^{\frac{3}{2}}}\\B=1.21*10^{-10}T

6 0
4 years ago
A 0.50 kg object is at rest. A 2.88 N force to the right acts on the object during a time interval of 1.48 s. a) What is the vel
Murljashka [212]

Answer:

8.5m/s

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We are given that

Mass of object=m=0.50 kg

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Force=F=2.88 N

Time=1.48 s

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2.88=0.50a

a=\frac{2.88}{0.50}=5.76m/s^2

a=\frac{v-u}{t}

Using the formula

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Can different masses have the same kinetic energy?
skelet666 [1.2K]
Yes.objects of different masses can have the same kinetic energy
8 0
3 years ago
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