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3 years ago

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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2 years ago

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2 years ago

What speed would a fly with the mass of 0.55 g need in order to have the same kinetic energy as the automobile in the term 19

larisa86 [58]

The question does not provide enough information to complete the answer, so I'll assume the needed data to help you to solve your own problem

Answer:

The fly should need to move at 9,534.6 m/s to have the same kinetic energy as the automobile

Explanation:

Kinetic Energy

Is the capacity of a body to do work due to its speed and is computed by

$\displaystyle K=\frac{mv^2}{2}$

We are not given enough data to compare the kinetic energy of the fly with that of the automobile. We'll assume the following characteristics:

$m_a=500\ kg$

$v_a=10\ m/s$

So its kinetic energy is

$\displaystyle K_a=\frac{(500)10^2}{2}$

$\displaystyle K_a=25,000\ J$

The mass of the fly is

$m_f=0.55\ gr=0.00055\ kg$

To have the same kinetic as the automobile:

$\displaystyle \frac{m_fv_f^2}{2}=25,000$

Solving for $v_f$

$\displaystyle v_f=\sqrt{\frac{2(25,000))}{m_f}}$

$\displaystyle v_f=\sqrt{\frac{50,000}{0.00055}}$

$v_f=9,534.6\ m/s$

The fly should need to move at 9,534.6 m/s to have the same kinetic energy as the automobile

5 0

3 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0

3 years ago

Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ = 4.0 × 10-6 C

Len [333]

Answer: 71.93 *10^3 N/C

Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:

∫E*dr=Q inside/εo Q inside is given by: λ*L then,

E*2*π*r*L=λ*L/εo

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6 0

3 years ago

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

How many atoms are in a 4.7 g copper coin?