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Liono4ka [1.6K]
3 years ago
13

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

Physics
1 answer:
denpristay [2]3 years ago
8 0

Answer:

Approximately 5.5\; \rm N, assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let q_1 and q_2 denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, q_1 = 20 \times 10^{-6}\; \rm C  and q_2 = 15 \times 10^{-6}\; \rm C.

Let r denote the distance between these two point charges. In this question, r = 0.7\; \rm m.

Let k denote the Coulomb constant. In standard units, k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}.

Substitute in the values and evaluate:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}.

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Explanation:

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7 0
3 years ago
The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o
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Answer:

F_f=840N

Explanation:

From the question we are told that

Weight of fireman W_f= 87.5kg

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Final speed is V_f 1.75m/s

Generally the equation for velocity is mathematically represented as

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Generally the equation for Frictional force F_f is mathematically given as

F_f=m*a

F_f=m*(g-0.21)

F_f=87.5*(9.81-0.21)

Therefore

F_f=840N

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THEN next you get your Balmer series, where the electrons fall in
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3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
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