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IrinaVladis [17]
1 year ago
12

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 th

e lifetime of a top quark)? Note that these powers of 10 follow the same rules that any exponents would follow.
(The age of the universe is around 100,000,000,000,000,000s. A top quark has a lifetime of roughly 0.000000000000000000000001s. Writing 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.)
Physics
1 answer:
ololo11 [35]1 year ago
6 0

Answer:

1.0\cdot 10^{41} times

Explanation:

First of all, we need to write both the age of the universe and the lifetime of the top quark in scientific notation.

Age of the universe:

T=100,000,000,000,000,000s = 1.0\cdot 10^{17} s (1 followed by 17 zeroes)

Lifetime of the top quark:

\tau = 0.000000000000000000000001s = 1.0\cdot 10^{-24} s (we moved the decimal point 24 places to the right)

Therefore, to answer the question, we have to calculate the ratio between the age of the universe and the lifetime of the top quark:

r = \frac{T}{\tau}=\frac{1.0\cdot 10^{17} s}{1.0\cdot 10^{-24} s}=1.0\cdot 10^{41}

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A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li
babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

y=\frac{m\lambda D}{d}

At m = 0

y_0=0

y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm

At m = 1

y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m

The slit separation is 0.00001266 m

3 0
2 years ago
A child on a playground swing is swinging back and forth (one complete oscillation) once every four seconds, as seen by her fath
Sati [7]

Answer:

A

Explanation:

Solution:-

- According to the law of relativity the relative speed between two moving objects is inversely proportional to the the time taken.

- Ignoring Doppler Effect.

- So if the relative speeds of two objects in motion i.e ( swing and spaceship) are positive then the time frame of reference for both object relative to other other decreases. So in other words if spaceship approaches the swing i.e relative velocity is positive then the time period of oscillation observed would be less than actual i.e less than 4 seconds.

- Similarly, if spaceship moves away from the swing i.e relative velocity is negative then the time period of oscillation observed would be more than actual i.e more than 4 seconds.

4 0
1 year ago
Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is
topjm [15]

Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

4 0
1 year ago
An airliner of mass 1.70×105kg1.70×105kg lands at a speed of 75.0 m/sm/s. As it travels along the runway, the combined effects o
Karo-lina-s [1.5K]

Answer:

The airliner travels 1.65 km along the runway before coming to a halt.

Explanation:

Given

Resistive forces = (2.90 × 10⁵) N = 290000 N

Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg

Velocity of airliner = 75 m/s

Let the distance over moved by the airliner be equal to d

According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.

Work done by the resistive forces = (290000) × d = (290,000d) J

Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J

290000d = 478,125,000

d = (478,125,000/290,000)

d = 1648.7 m = 1.65 km

Hope this helps!!!

4 0
2 years ago
A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 1010 excess electrons
liraira [26]

When the metals touch together, half the charge of the charged metal flows to the other because the electrons all repel each other. Therefore this also means that each metal ball contains the same amount of electrons. Each ball has 5^10 electrons, this is equivalent to a total charge of:

Q1 = Q2 = (1.602 * 10^-19 coulombs / electron) 5^10 electrons = Q

Q = 1.564 * 10^-12 C

 

Now using the Coulombs law to find for the electric force:

F = k q1 q2 / r^2 = k (Q)^2 / r^2

where k is a contant = 9 * 10^9 N m^2 / C^2

r = the distance of the two metals = 0.2 m

So,

F = (9 * 10^9 N m^2 / C^2) (1.564 * 10^-12 C)^2 / (0.2 m)^2

F = 5.51 * 10^-13 N

 

Since the two metals repel therefore they are the one which exerts the force hence the magnitude must be negative:

<span>F = - 5.51 * 10^-13 N</span>

7 0
2 years ago
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