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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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A 1.0 kg ball falls from rest a distance of 19.6 m.
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Answer:

192.08J

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Explanation:

Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.

PE=mgh

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=192.08J

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v²=u²+2as

v²=0²+2(9.8)(19.6)

v=√384.16

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6 0
3 years ago
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kati45 [8]

Answer:

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6 0
3 years ago
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Suppose a free-fall ride at an amusement park starts at rest and is in free fall. What is the velocity of the ride after 2.3 s?
horsena [70]

Answer:

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Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

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\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

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w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

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