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dedylja [7]
3 years ago
13

A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th

en what is the speed of ball
Physics
1 answer:
ANEK [815]3 years ago
5 0

Answer:\dfrac{\pi}{2} ms^{-1}

Explanation:

Let T be the time required to make one revolution.

Let r be the radius of the circular path.

Let d be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So,d=2\pi r

Given,d=0.5m\\T=2sec

d=2\times \pi \times 0.5=\pi m

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let s be the speed of the ball.

s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}

So,the speed of the ball is \frac{\pi }{2}ms^{-1}

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Answer:

The value is E_i  =  1.5596 *10^{-18} \  J

Explanation:

From the question we are told that

The wavelength is \lambda  =  48.2 nm  =  48.2  *10^{- 9 }\  m

The velocity is v = 2.371*10^6 \ m/s

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Generally the energy of the incident light is mathematically represented as

E =  \frac{h *  c}{\lambda}

Here c is the speed of light with value c =  3.0 *10^{8} \  m/s

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So

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=> E = 4.12 *10^{-18} \  J

Generally the kinetic energy is mathematically represented as

E_k  =  \frac{1}{2} *  m_e * v^2

=> E_k  =  \frac{1}{2} *  9.109*10^{-31} * (2.371*10^6 )^2

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Generally the ionization energy is mathematically represented as

E_i  =  4.12 *10^{-18} -   2.56 *0^{-18}

=>     E_i  =  1.5596 *10^{-18} \  J

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