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grigory [225]
3 years ago
9

a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second

than there are seconds in a year? (b) Humans have existed for about 106 years, whereas the universe is about 1010 years old. If the age of the universe is defined as 1 "universe day," where a universe day consists of "universe seconds" as a normal day consists of normal seconds, how many universe seconds have humans existe
Physics
1 answer:
lawyer [7]3 years ago
3 0

Answer:

a) Yes, there are 10^8 shakes in a second (1 s \frac{1 shake}{10^{-8}s}=10^8 shake) in a year there are 31,536,000 seconds... that is 3.1536x10^7 (1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7)

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s

That is about 2 and half hours.

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Total mechanical energy (The sum of kinetic and potential energy
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Answer:

Emechanical=mgh+\frac{1}{2}mν²

Explanation:

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Emechanical=Epotential+Ekinetic

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5 0
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The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t
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Answer:

T₂ = 123.9 N,  θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

       T₁ₓ - T_{2x} = 0

        T_{2x}= T₁ₓ

Y axis y

      T_{1y} + T_{2y} - 200N = 0

      T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

         sin 60 = T_{1y} / T₁

         cos 60 = t₁ₓ / T₁

         T_{1y} = T₁ sin 60

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for voltage 2 it is done in the same way

         T_{2y} = T₂ sin θ

         T₂ₓ = T₂ cos θ

we substitute

         

           T₂ sin θ= 200 - 86.6 = 113.4

           T₂ cos θ = 50              (1)

to solve the system we divide the two equations

           tan θ = 113.4 / 50

           θ = tan⁻¹ 2,268

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we caption in equation 1

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           T₂ = 50 / cos 66.2

           T₂ = 123.9 N

8 0
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