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3 years ago

The magnitude of the charge of the electron is:

Physics

1 answer:

brilliants [131]3 years ago

3 0

Answer:

a. Exactly the same as the magnitude of the charge of the proton.

Explanation:

The elementary charge (e) is the smallest electric charge that can exist in the universe. Any positive or negative electric charge can be expressed as a multiple of the elementary charge, since is the electric charge carried by a single proton or, equivalently, the magnitude of the electric charge carried by a single electron (-1e).

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Que cuerpos celestes observó laika durante su viaje

sergejj [24]

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3 years ago

An object is in circular motion. How will the object behave if the centripetal force is removed

stich3 [128]

The object will sail away in a straight line ... continuing in the same direction it was going when the centripetal force stopped.

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3 years ago

Emily and Gemma did a Reaction time lab. Emily dropped the ruler while Gemma tried to catch it. She caught the ruler 5 times and

dlinn [17]

The average reaction time of Gemma is 0.1564 seconds.

As we know, Gemma is catching the scale and Emily is dropping the scale.

The whole experiment is taking place under gravity, so the acceleration is constant.

As we know, the scale is dropped, it means that the initial velocity of the scale is zero.

We can use the equation of motion,

The equation is,

S = Ut + 1/2at²

Where,

S is the displacement, which is 0.12 m in our case,

U is initial velocity which is 0m/s because the stone is dropped,

t is the time taken, this is equal tot he reaction time here,

a is the acceleration due to gravity whose value is 9.8m/s.

Now, putting all the values,

0.12 = 1/2(9.8)(t)²

t² = 0.24/9.8

t = 0.1564 seconds.

Gemma reacts in 0.1564 seconds to catch the scale.

To know more about equation of motion, visit,

brainly.com/question/27821888

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1 year ago

Which character trait below is a trait that both Michael Faraday and Nikola Tesla had?

-Dominant- [34]

They were both odd and obsessive

7 0

3 years ago

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res

garri49 [273]

Answer:

ω = 22.36 Hz

f = 3.56 Hz

T = 0.28 s.

Explanation:

a) The angular frequency (ω), can be calculated using the following equation:

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{F}{x*m}}$

Where:

k: is the spring constant = F/x

m: is the mass of the particle = 0.500 kg

F: is the force applied = 7.50 N

x: is the displacement = 3.00 cm = 0.03 m

$\omega = \sqrt{\frac{7.50 N}{0.03 m*0.500 kg}} = 22.36 s^{-1} = 22.36 Hz$

Therefore, the angular frequency of the motion is 22.36 Hz.

b) To find the frequency (f) we can use the next equation:

$f = \frac{\omega}{2 \pi} = \frac{22.36 Hz}{2 \pi} = 3.56 Hz$

Hence, the frequency of the motion is 3.56 Hz.

c) The period (T) is equal to:

$T = \frac{1}{f} = \frac{1}{3.56 Hz} = 0.28 s$

Therefore, the period of the motion is 0.28 s.

I hope it helps you!

4 0

3 years ago