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3 years ago

How many electrons are there in 3.5 x 10" C?

Physics

1 answer:

Masja [62]3 years ago

4 0

Answer:

$2.19\cdot 10^{20}$ electrons

Explanation:

I assume you mean:

How many electrons are there in $3.5\cdot 10^{1}C$ ?

Solution:

The charge of one electron is (in magnitude)

$e=1.6\cdot 10^{-19}C$

The charge in this problem is

$Q=3.5\cdot 10^{1} C$

So, we can find how many electrons are in this charge by simply dividing the total charge by the charge of one electron:

$n=\frac{Q}{e}=\frac{3.5\cdot 10^{1}}{1.6\cdot 10^{-19}}=2.19\cdot 10^{20}$

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1 year ago

The wheelchair starts from rest. It accelerates at a constant rate until it has a speed of 1.5 m/s. The wheelchair travels a dis

anastassius [24]

<h2>Answer:-</h2>

$\pink{\bigstar}$ The acceleration of the wheelchair is $\large\leadsto\boxed{\tt\purple{0.5625 \: m/s^2}}$

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<h3>• Given:-</h3>

Initial velocity of the wheelchair = 0 m/s

Final velocity of the wheelchair = 1.5 m/s

Distance covered by the wheelchair = 2 m

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Acceleration of the wheelchair = ?

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<h3>• Solution:-</h3>

We know,

• Third Equation of Motion:-

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{v^2 - u^2 = 2as}}}$

where,

u = Initial velocity

v = Final velocity

a = Acceleration

s = Distance covered

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Substituting the values in the Formula:-

➪ $\sf (1.5)^2 - (0)^2 = 2 \times a \times 2$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf 2.25 - 0 = 4 \times a$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf 2.25 = 4 \times a$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf a = \dfrac{2.25}{4}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ $\large{\bold\red{a = 0.5625 \: m/s^2}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, the acceleration of the wheelchair is 0.5625 m/s².

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3 years ago

An iron nail floats in mercury and sinks in water. explain why?

denpristay [2]

Answer:

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3 years ago

Draw a wave that has a wavelength of 3 cm and an amplitude of 1 cm. Label the wavelength, the amplitude, the rest position, and

Minchanka [31]

Answer:

Please find attached, the required wave drawn with MS Excel

Explanation:

Functions that represent waves is given as follows

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Where;

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At constant v, λ ∝ 1/f

∴ λ ∝ T

Where T = 3, we have;

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∴ B = 2·π/3

Therefore, we have the wave with an amplitude of 1 cm, and wavelength, 3 cm, given as follows

y = sin((2·π/3)·x)

Plotting the above wave with MS Excel, we can get the attached wave

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3 years ago