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Akimi4 [234]
3 years ago
10

What do electrons in the same shell have in common? AThey have the same amount of energy incorrect answer BThey are all positive

ly charged incorrect answer CThey are all made up of atoms incorrect answer DThey all have neutral charges incorrect answer
Physics
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

They have the same amount of energy

Explanation:

Electrons are said to be the subatomic particles that move around the nucleus of an atom. These electrons are negatively charged particles that are seen to be quite smaller than the nucleus of an atom.  

The electron shells of these atoms are usually being filled from the inside out with the low-energy shells closer to the nucleus being filled before they can go into the much higher-energy shells that are a bit out

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Here we can use the work energy theorem

W_f + W_s = K_f - K_i

here we know that

K_f = 0

as it come to rest finally

K_i = \frac{1}{2}mv_i^2

K_i = \frac{1}{2}\times 1\times 2^2

K_i = 2 J

now work done by friction force will be given as

W_f = - F_f \times d = -\mu mg d

W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu

Work done by spring force is given as

W_s = \frac{1}{2}k(x_i^2 - x_f^2)

W_s = \frac{1}{2}(10)( 0 - 0.10^2)

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so now plug in all data above

- 0.05 - \mu(0.98) = 0 - 2

\mu = 1.99

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4 years ago
Why is it important to know how fast a roller coaster travels at certain points on the track?
jolli1 [7]

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6 0
3 years ago
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A particular field is 111 yards long. Express this length in meters.
romanna [79]

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Gravity prevents planets from flying off at a
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A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

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The distance in meters traveled by a point outside the rim is calculated as follows;

\theta = \omega t\\\\\theta = (100 \frac{rev}{\min}  \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi  \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

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3 years ago
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