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Gemiola [76]
3 years ago
7

How many atoms are in a 4.7 g copper coin?​

Physics
2 answers:
azamat3 years ago
4 0
3.11 is the answer I think
WINSTONCH [101]3 years ago
3 0

Answer:

x = 4.45 * 10 ^22  Note. Technically, this should be rounded to 4.5 * 10^22. There are only 2 sig digits.

Explanation:

You have to assume that the coin is pure copper, which I doubt.  What a coin is actually made of  depends on when it was minted. But for the sake of this question, we'll assume coins are pure copper.

Copper has an atomic mass of 63.546 grams / mol

So 4.7 g of copper = 4.7 / 63.545 mol

We have 0.07396 mol of copper

1 mol of anything = 6.02 * 10^23 atoms (in this case).

0.07396 mol        = x

Cross Multiply

               

1 * x = 0.07396 * 6.02 * 10^23

x = 4.45 * 10 ^22  atoms of copper

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Which of the following statements is/are true?Check all that apply.a. A conservative force permits a two-way conversion between
Eddi Din [679]

Answer:

a). A conservative force permits a two-way conversion between kinetic and potential energies.

TRUE

Because there is no energy loss in presence of conservative forces so energy conversion in two ways are possible.

b). A potential energy function can be specified for a conservative force.

TRUE

negative gradient of potential energy is equal to conservative force

F = -\frac{dU}{dr}

c). A non-conservative force permits a two-way conversion between kinetic and potential energies.

FALSE

here energy is lost against non-conservative forces

d). The work done by a conservative force depends on the path taken.

FALSE

work done by conservative force is independent of path

e). The work done by a non-conservative force depends on the path taken.

TRUE

work done by non conservative forces depends on path.

f). A potential energy function can be specified for a non-conservative force.

FALSE

It is not defined for non conservative forces

3 0
3 years ago
Rays of light coming from the sun (a very distant object) are near and parallel to the principal axis of a concave mirror. After
Setler [38]

A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.

A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror.  This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.

In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.

Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.

For reference: brainly.com/question/20380620

3 0
2 years ago
Which of these actions will increase friction? Select three options. Scratching a surface to make it rougher polishing a surface
k0ka [10]

Answer:

scratching a surface to make it rougher

increasing the size of a flying object

adding extra weight to an object

Explanation:

5 0
2 years ago
Read 2 more answers
What is your acceleration while sitting in your chair. the latitude of corvallis is 44.4˚.?
marta [7]
 <span>You can start with the equations you know 

a=v^2/r = (2pi*r/T)^2/r = 4pi^2r/T^2 

Radius of earth (R) = 6378.1 km 
Time in one day (T) = 86400 seconds 
Latitude = 44.4 degrees 

If you draw a circle and have the radius going out at a 44.4 degree angle above the center you can then find the r. 

r=Rcos(44.4) 
r=6378.1cos(44.4) 
r= 4556.978198 km or 4556978 m 

Now you can plug this value into the acceleration equation from above... 

a= 1.8*10^8/7.47*10^9 
a= .0241 m/s^2 </span>
8 0
2 years ago
A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

5 0
3 years ago
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