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Leviafan [203]
3 years ago
7

Roughly speaking, the radius of an atom is about 10,000 times greater than that of its nucleus. If an atom were magnified so tha

t the radius of its nucleus became 2.0 cm, about the size of a marble, what would be the radius of the atom in miles
Physics
1 answer:
skad [1K]3 years ago
4 0

Answer:

0.124 miles

Explanation:

Since the radius of the atom R = 10000r where r = radius of nucleus. Now, if r = 2.0 cm = 0.02 m,

R = 10000r

= 10000 × 0.02 m

= 200 m

We know that 1 mile = 1609 m.

So the radius of the atom in miles is R = 200 m × 1 mile/ 1609 m = 0.124 miles

So, the radius of the atom in miles is 0.124 miles  

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If the spring constant is doubled, what value does the period have for a mass on a spring? A. The period would double by square
antoniya [11.8K]

Answer:

The period would decrease by sqrt(2)

Explanation:

The restoring force is given by,

F = -kx

According to Newton's second law of motion,

ma = -kx

ma + kx = 0

The time period is given by,

T =\frac{2\pi }{\omega }

Where \omega is the angular velocity and it is given by,

\omega = \sqrt{\frac{k}{m} }

Now if the spring constant is doubled then,

k_{2} = 2k

Thus,

T_{2} =\frac{2\pi }{\sqrt{\frac{2k}{m} } }

\frac{T_{2} }{T} = \frac{\frac{2\pi }{\sqrt{\frac{2k}{m} } }}{\frac{2\pi }{\sqrt{\frac{k}{m} } }}

\frac{T_{2} }{T} = \sqrt{\frac{k}{2k} } = \sqrt{\frac{1}{2} }

T_{2} = \frac{T}{\sqrt{2} }

Thus, The period would decrease by sqrt(2).

Hence, option D is correct.

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3 years ago
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xxTIMURxx [149]

Answer:

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Explanation:

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