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Nostrana [21]
2 years ago
10

A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change

the period to 2.00 s
Physics
1 answer:
Daniel [21]2 years ago
7 0

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

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

Where K indicates spring constant

m indicates mass

For the new time period

T^' = 2\pi \sqrt{\frac{m'}{k} }

Now, we will take 2 ratios of the time period

\frac{T}{T'} = \sqrt{\frac{m}{m'} }

\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }

0.5625 = \sqrt{\frac{0.500}{m'} }

m' = \frac{0.500}{0.5625}

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

or

= 389 kg

Therefore for computing the mass we simply applied the above formula.

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5 0
3 years ago
A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.
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Answer:

v=33.764m/s

Explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

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Velocity v

Solution

From the Centripetal force we know that

F_{c}=\frac{mv^{2} }{r}

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

So

F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s

4 0
2 years ago
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