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BartSMP [9]
3 years ago
11

A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of at

oms are there in the sheet?
Physics
1 answer:
Olegator [25]3 years ago
7 0

Answer:

719

Explanation:

Conversion

1 picometer (pm) is equivalent to 1 × 10^{-12} meter

1 micrometer is equivalent to 1 × 10^{-6} meter

To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by 174 × 10^{-12} meter we get that the number of sheets will be as follows

\frac {0.125× 10^{-6}}{174\times 10^{-12}}=718.3908045\approx 719

Therefore, they are approximately 719 sheets

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A wave period is the time it takes to complete one cycle


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A 70.0-kilogram man is walking at a speed if 2.0 m/s. What is the kinetic energy?
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Answer:

140 J

Explanation:

From the the question, the mass of the man =70.0 kg and the speed at which the man is walking =2.0 m/s.

K.E =  \frac{1}{2}m {v}^{2}

where K.E = the kinetic energy, m=mass and v= speed.

By substitution,

K.E =  \frac{1}{2} \times 70 \times  {2}^{2}

\implies \: K.E = \frac{280}{2}

\implies K.E =140

Hence the kinetic energy of the man is 140J

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Which of the following are metric units? gallons kilograms pounds inches
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kilogram

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3 years ago
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Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side
pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m  + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

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