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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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Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

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