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wel
3 years ago
13

A gold bar contains 63.0 mol of gold, Au(s). How many atoms of gold are in the bar?

Physics
1 answer:
vredina [299]3 years ago
7 0

Answer:

63.0 x (6.02 x 10^23) = 3.79 x 10^25 atoms of gold

Explanation:

There are 6.02 x 10^23 representative particles in a mole of any substance. Multiply it by the number of moles, and you will get the number of representative particles (atoms, molecules, formula units)

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The sound produced by the loudspeaker in the drawing has a frequency of 11999 Hz and arrives at the microphone via two different
const2013 [10]

The speed at which sound travels through the gas in the tube is 719.94m/s

<u>Explanation:</u>

Given:

Frequency, f = 11999Hz

Wavelength, λ = 0.03m

Velocity, v = ?

Sound speed in the tube is calculated by multiplying the frequency v by the wavelength λ.

As the sound loudness changed from a maximum to a minimum, then we know the sound interference in the case changed from constructive interference (the two sound waves are in phase, i.e. peaks are in a line with peaks and so the troughs), to a destructive interference (peaks coinciding with troughs). The least distance change required to cause such a change is a half wavelength distance, so:

λ/2 = 0.03/2

 λ  = 0.06m

We know,

v = λf

v = 0.06 X 11999Hz

v = 719.94m/s

Therefore, the speed at which sound travels through the gas in the tube is 719.94m/s

3 0
3 years ago
What disadvantage of convex mirror​
Eddi Din [679]

Answer:

The disadvantage is that convex mirror is that they make it appear like those objects are at a more noteworthy distance than they actually are. It always gives a virtual, erect and a diminished image.

Explanation:

5 0
3 years ago
the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path
Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:  

V=\sqrt{G\frac{M}{r}}

Where:  

V is the speed of travel of the Moon around the Earth

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24} kg is the mass of the Earth

r=384400(10)^{3} m is the distance from the center of the Earth to the center of the Moon

Solving:

V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}

V=1018.26 m/s This is the speed of travel of the Moon around the Earth

5 0
3 years ago
1. A large turbine has an initial angular momentum of 6700 kgm^2/s. A storm is rolling in and the wind picks up. 8 seconds later
LekaFEV [45]

Answer:

262.5 Nm

Explanation:

Torque is the rate of change of angular momentum.

Hence, we have

\tau = \dfrac{\Delta L}{t}

Δ<em>L</em> is the change in angular momentum.

Using values in the question,

\tau = \dfrac{8800-6700 \text{ kg m}^2\text{/s}}{8\text{ s}}  = 262.5 \text{ Nm}

4 0
3 years ago
Leonard is stressed and feeling overwhelmed by the college selection process. What should he do to combat this problem?
exis [7]
He should ask other people for advice
6 0
3 years ago
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