Scientific knowledge builds upon previous knowledge."

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

Olegator [25]

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

7 0

2 years ago

A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp

Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

$y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}$

1. Finding Ф from the initial conditions:

$-0.02 = 0.02cos(0 + \phi) => \phi = \pi$

2. Finding time t at position y = 1 cm:

$0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}$

3. Find velocity v at time t from equation 2:

$v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))$

5 0

2 years ago

Read 2 more answers

A 10-meter long ramp has a mechanical advantage of 5. What is the height of the ramp?

denpristay [2]

<span><span>1. </span>If the ramp has a length of 10 and has a mechanical advantage (MA) of 5. Then we need to find the height of the ramp.
Formula:
MA = L / H
Since we already have the mechanical advantage and length, this time we need to find the height .
MA 5 = 10 / h
h = 10 / 5
h = 2 meters

Therefore, the ramp has a length of 10 meters, a height of 2 meters with a mechanical advantage of 5.</span>

6 0

2 years ago

Consider a string of length 1.0 meter, fixed at both ends, with mass 100 grams and tension 100 newtons. part a give the number o

Bond [772]

To answer the problem we would be using this formula which isv = sqrt(T/(m/L))
v = sqrt(100 N / [(0.100 kg)/(1.0 m)])
v = 31.62 m/s
v = fλ
31.62 m/s = (95 Hz)(λ)
λ = 0.333 m
For every wavelength along a string there will be 2 antinodes.
1.0 m / 0.333 m = 3
3 * 2 = 6 antinodes
6 + 1 = 7 nodes

4 0

2 years ago

A tennis ball of mass m=0.060 kg and speed v=25 m/s strikes a wall at a 45 angle and rebounds with the same velocity at 45°. Wha

Diano4ka-milaya [45]

To solve this problem we will apply the concepts related to the Impulse which can be defined as the product between mass and the total change in velocity. That is to say

$p = m\Delta v$