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garri49 [273]
3 years ago
13

Pennies today are copper-covered zinc, but older pennies are 3.1 g of solid copper. What are the total positive charge and total

negative charge in a solid copper penny that is electrically neutral?
Physics
1 answer:
k0ka [10]3 years ago
7 0

Answer:

Q_+=1.36*10^5C

Q_+=-1.36*10^5C

Explanation:

From the question we are told that

Mass of old penny M=3.1g

Generally the number of moles in a penny is given mathematically as

n=\frac{M}{A}

A=atomic\ number\ of\ copper=>63.5g/mol

Therefore

n=\frac{3.1}{63.5}

n=0.049mol

Avogadro\ Number=(6.023*10^2^3 atom/mol

Therefore

N=(6.023*10^2^3 atom/mol *0.049mol

N=2.939*10^2^2atoms

Generally the Total positive charge of the copper is given by

Since its the 29th atom of the periodic table

29 protons

29 electrons

Q_+=29(2.939*10^22)(1.6*10^1^9)

Q_+=1.36*10^5C

Generally the Total negative charge of the copper is given by

Q_+=29(2.939*10^22)(-1.6*10^1^9)

Q_+=-1.36*10^5C

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Two iron bolts of equal Mass one at a hundred see another at 55 Sierra place in the insulated cylinder assuming the heat capacit
malfutka [58]

Answer:

T_2 = 77.5c

Explanation:

From the question we are told that

Temp of first boltsT_1=100

Temp of 2nd bolt T_2=55

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  q=cm \triangle T

Generally heat released  by the iron bolt  = heat gained by the iron bolt

Generally solving mathematically

     -(0.45*m* (T_2-100  \textdegree c)) = 0.45*m*(T_2 -55\textdegree c)

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5 0
2 years ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

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Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
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lozanna [386]

Answer:

Explanation:

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6 0
3 years ago
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