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dezoksy [38]
3 years ago
15

A 95 kg clock initially at rest on a horizontal floor

Physics
1 answer:
Nataly [62]3 years ago
4 0

Answer:

You are given that the mass of the clock M is 95 kg.

This is true whether the clock is in motion or not.

Fs is the frictional force required to keep the clock from moving.

Thus Fk = uk W = uk M g      the force required to move clock at constant speed.     (the kinetic frictional force)

uk = 560 N / 931 N = .644   since the weight of the clock is 931 N  (95 * 9.8)

us  is the frictional force requited to start the clock moving

us = static frictional force = 650 / 931 -= .698

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Structures on a bird feather act like a diffraction grating having 8000 8000 lines per centimeter. What is the angle of the firs
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Answer:

Angle of first order maximum, \theta=21.19^{\circ}

Explanation:

Given that,

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Convert 6.6 meters to centimeters
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Answer:

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There are 100 cm in 1 m. To convert from m to cm, multiply by 100.

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A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact
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Answer:

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Explanation:

The equation for radioactive decay its:

N ( t) \ = \ N_0 \ e^{ \ -  \frac{t}{\tau}},

where N(t) its quantity of material at time t, N_0 its the initial quantity of material and \tau its the mean lifetime of the radioactive element.

The half-life t_{\frac{1}{2}} its the time at which the quantity of material its the half of the initial value, so, we can find:

N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

so:

\ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}

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N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  t_{\frac{1}{2}}}{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  \tau \ ln( 2 ) }{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  10 \  \ ln( 2 ) }

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}

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10 \ * \ 24,110 \ years \ = \ 241,100 \ years

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

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