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ss7ja [257]
3 years ago
7

In a 100-m race, the winner is timed at 11.2 s. The second-place finisher’s time is 11.6 s. How far is the second-place finisher

behind the winner when she crosses the finish line? Assume the velocity of each runner is constant throughout the race.
Physics
1 answer:
Veronika [31]3 years ago
3 0

Answer:

d = 3.45 m

Explanation:

given,

distance, d = 100 m

time of winner = 11.2 s

time of finisher = 11.6 s

distance between them when winner finish race

speed of winner

s_1 = \dfrac{d}{t_1}

s_1 = \dfrac{100}{11.2}

 s₁ = 8.93 m/s

speed of runner up

s_2 = \dfrac{d}{t_2}

s_2= \dfrac{100}{11.6}

 s₂= 8.62 m/s

difference in time = t₂ - t₁ = 11.6 - 11.2 = 0.4 s

distance between them

d =  s₂ x t

d = 8.62 x 0.4

d = 3.45 m

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Answer:

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Explanation:

The equation for radioactive decay its:

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

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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}

-  \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )

t_{\frac{1}{2}}\ = \tau ln( 2 )

So, after 10 half-lives, we got:

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}

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

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

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

7 0
3 years ago
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2 years ago
How are fission and fusion similar?
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Answer:

C

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8 0
2 years ago
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m2. The wave is incident u
Hatshy [7]

Answer:

The  total energy is  T  =  169.02 \ J

Explanation:

From the question we are told that

    The  Poynting vector (energy flux ) is  k  =  0.939 \ W/m^2

    The length of the rectangle is  l  =  1.5  \ m

    The  width of the rectangle is  w =  2.0 \ m

    The time taken is t  =   1 \ minute  =  60 \ s

The total electromagnetic energy falls on the area is mathematically represented as

      T  =  k  *  A  *  t

Where A  is the area of the rectangle which is mathematically represented as

           A= l *  w

 substituting values

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substituting values

        T  =  0.939 *  3 *  60

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5 0
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Answer:

(d) 2840 A

Explanation:

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(λ)= hc/ Φ

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But (λ) sodium is given as 5460 A,

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If we substitute the values we can find

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(λ)tungsten= [(λ) sodium × Φsodium]/ Φtungsten

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= 2854A

Hence, threshold wavelength of tungsten is 2854A

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