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vovikov84 [41]
3 years ago
7

The age of the universe is around 100,000,000,000,000,000s. A top quark has a lifetime of roughly 0.000000000000000000000001s. W

riting numbers out with all these zeros is not very convenient. Such quantities are usually written as powers of 10. The age of the universe can be written as 1017s and the lifetime of a top quark as 10−24s.Note that1017 means the number you would get by multiplying 10 times 10 times 10... a total of 17 times. This number, as you can see above, would be a one followed by seventeen zeros. Similarly, 10−24 is the result of multiplying 0.1 (or 1/10) times itself 24 times. As seen above, this is written as 23 zeros after the decimal point followed by a one.***How many top quark lifetimes have there been in the history of the universe (i.e., what is the age of the universe divided by the lifetime of a top quark)? Note that these powers of 10 follow the same rules that any exponents would follow.
Physics
1 answer:
QveST [7]3 years ago
7 0

Answer:

10^{41}

Explanation:

Age of the universe = 100,000,000,000,000,000=10^{17}\ s

Lifetime of a top quark = 0.000000000000000000000001=10^{-24}

Top quark life would be given by

t=\dfrac{\text{Age of the universe}}{\text{Lifetime of a top quark}}\\\Rightarrow t=\dfrac{10^{17}}{10^{-24}}\\\Rightarrow t=10^{17+24}\\\Rightarrow t=10^{41}\ s

Hence, the answer is 10^{41}\ s

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Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le
Nana76 [90]

Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = \frac{d}{2}

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s

Energy

E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J

The new energy will be 1000 J

Work done will be the change in the kinetic energy

W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J

The work done is 600 J

5 0
3 years ago
Drag each tile to the correct location.
In-s [12.5K]

Answer:

Look at the image please

Explanation:

5 0
3 years ago
Write any two features of capital​
amid [387]
Washington DC and new Mexico
6 0
3 years ago
Read 2 more answers
Does the moon actually change shape as it revolves/rotates around the Earth? Explain your answer
rusak2 [61]

<u>Answer:</u>

<em>The moon doesn’t change shape on its own.</em>

<u>Explanation:</u>

Shapes of moon that we observe is based on the different perspectives of view from the earth and position of moon with respect to the sun. The changes arise due to the rotation of earth on its own axis as well as the revolution of moon on its orbit. The moon doesn’t have any light of its own.

It just reflects off the light from the sun. Due to tidal locking phenomenon one face of the moon permanently faces the sun. Because of the changes in position of moon with respect to the sun the moon is lighted up variably giving rise to various phases like new moon, full moon, crescent etc.

3 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
Sergeeva-Olga [200]

Answer:

33.23 m

Explanation:

At the point where both objects will meet, the vertical height will be equal.

From the equations of motion, the vertical height of the body falling at any time is given as

(y - y₀) = ut + ½gt²

y = vertical height at any time T

y₀ = initial height of the object = 81.5 m

u = initial velocity = 0 m/s (body was dropped)

g = -9.8 m/s²

(y - 81.5) = 0 - 4.9T²

y = 81.5 - 4.9T² (eqn 1)

For an object thrown up, the vertical height of the body at any time, t, is given as

(y - y₀) = ut + ½gt²

y = vertical height of the object at any time t

y₀ = initial height of the object = 0 m

u = initial velocity = 40 m/s

g = -9.8 m/s²

y = 40t - 4.9t² (eqn 2)

At the point where the two objects meet, we equate eqn 1 and eqn 2

y = y

81.5 - 4.9T² = 40t - 4.9t²

But T = (t + 2.2) (Since object 2 was dropped 2.2 s after object 1)

81.5 - 4.9(t + 2.2)² = 40t - 4.9t²

81.5 - 4.9(t² + 4.4t + 4.84) = 40t - 4.9t²

81.5 - 4.9t² - 21.56t - 23.716 = 40t - 4.9t²

81.5 - 21.56t - 23.716 - 40t = 0

57.784 = 61.56t

t = (57.784/61.56) = 0.93866 = 0.94 s

Therefore, the vertical height at t = 0.93866 s is

y = (40×0.93866) - 4.9(0.93866²) = 33.23 m

Hope this Helps!!!

5 0
3 years ago
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