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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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8 0
3 years ago
A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel
erastovalidia [21]

Answer:

D.

Explanation:

To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by

2t = (m+\frac{1}{2})\frac{\lambda}{n}

Where,

t = thickness

\lambda=Wavelenght

m= is an integer

n= film/refractive index

We use this equaton because phase change is only present for gasoline air interface, but not at the gasoline-water interface. <em>The minimum t only would be when the value of m=0 then</em>

2nt = \frac{\lambda}{2}

t = \frac{560nm}{4*1-4}

t = 100nm

Therefore the correct answer is D. The minimum thickness of the film to see ab right reflection is 100nm

4 0
3 years ago
Rocket-powered sleds are been used to test the responses of humans to acceleration. Starting from rest, one sled can reach a spe
Greeley [361]

Answer:cho  v₀ =0s  

α=Δv/Δt

Explanation:

\frac{0-495}{2,16-1,78}

=-1302,631579

chuyển động chậm dầnđều

3 0
2 years ago
A object weighing 5 kg, starts to accelerate evenly on a horizontal line. A force moves the object
hammer [34]
Work= force*distance
Work= x*12
Force= mass*acceleration
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5 0
4 years ago
A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an
posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle \theta=30°

We need to calculate the angle in radian

\theta=30\times\dfrac{\pi}{180}

\theta=0.523\ rad

We need to calculate the path length

Using formula of path length

Path\ length =angle\times radius

Path\ length=0.523\times5.9

Path\ length =3.09\ m

(b). Angle = 30 rad

We need to calculate the path length

Path\ length=30\times5.9

Path\ length=177\ m

(c). Angle = 30 rev

We need to calculate the angle in rad

\theta=30\times2\pi

\theta=188.4\ rad

We need to calculate the path length

Path\ length=188.4\times5.9

Path\ length =1111.56\ m

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0
3 years ago
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