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4 years ago

The subject of the second part of messiah is:

Physics

1 answer:

Zielflug [23.3K]4 years ago

5 0

The second part is about <span>his suffering and the spread of his doctrine.</span>

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

3 years ago

Can you please help me

konstantin123 [22]

Answer:

I dont know sorryyyyyyy

4 0

3 years ago

In an open system such as a campfire matter can

eimsori [14]

In an open system such as a campfire, matter can lose particles, gain particles or exchange particles.

4 0

3 years ago

Choose all correct sentences Group of answer choices The power is maximum when the value of the impedance is greater than the va

Illusion [34]

Answer:

True b and c

Explanation:

In an RLC circuit the impedance is

$Z = \sqrt{[R^{2} + ( (wL)^{2} + (\frac{1}{wC})^{2} ] }$

examine the different phrases..

a) False. The maximum impedance is the value of the resistance

b) True. Resonance occurs when

(wL)² + (1 / wC)² = 0

w² = 1 / LC

c) True. In resonance the impedance is the resistive part and the power is maximum

d) False. In resonance the inductive and capacitive part cancel each other out

e) False. The impedance is always greater outside of resonance, but at the resonance point they are equal

6 0

3 years ago

A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location

amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

$f=-16.0 cm$ (the focal length is negative for a diverging lens)

$p=10.0 cm$ is the distance of the object from the lens

Solvign the equation for q, we find

$\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}$

$q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm$

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0

3 years ago