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3 years ago

Which law describes the interactions between charged particles when they are not in contact?

Physics

2 answers:

natulia [17]3 years ago

7 0

The answer would be the coulombs law.

mojhsa [17]3 years ago

7 0

Answer:

B) Coulomb's law

Explanation:

As per coulombs law if two point charges are placed at some distance from each other then the force of interaction between them is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 , q_2$ = two charges between which force of interaction is required

d = distance between two charges

$k = \frac{1}{4\pi \epsilon_0}$

Faraday law is used to find the induced EMF due to changing magnetic flux

Ohm's law is used to relate potential difference between the ends of a conductor with its resistance and its current

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I need help in my physics class and show me how it’s done

Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

$A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)$

Where | A | is the magnitude of the vector and $\alpha$ is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle $\alpha$ .

Vector A is in the 4th quadrant

So:

$A_x = 6\\\\A_y = -6.5$

So:

$|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846$

So:

$Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)$

$\alpha$ = -47.28 ° +360° = 313 °

$\alpha$ = 313 °

Option 4.

4 0

3 years ago

Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in

iVinArrow [24]

Answer:

The value is $P_2 = 40.54 \ psla$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 14\ psla$

The initial temperature is $T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K$

The final temperature is $T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K$

Generally the equation for adiabatic process is mathematically represented as

$PT^{\frac{\gamma}{1- \gamma} } = Constant$

=> $P_1T_1^{\frac{\gamma}{1- \gamma} } = P_2T_2^{\frac{\gamma}{1- \gamma} }$

Generally for a monoatomic gas $\gamma = \frac{5}{3}$

$14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }$

=> $14 * 283^{-2.5} =P_2 * 433^{-2.5}$

=> $P_2 = 40.54 \ psla$

8 0

2 years ago

Evaluate (x +y)0 for x= -3 and y=5.<br> 0 1<br> 2<br> 01

frosja888 [35]

Answer:2.01201

Explanation:

8 0

3 years ago

Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger t

podryga [215]

Answer:

8F_i = 3F_f

Explanation:

When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.

Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.

Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.

The electrostatic force, Fi, in the initial configuration can be calculated as follows.

$F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f$

7 0

3 years ago

How much energy is needed to generate 0.71 x 10-16 kg of mass?

AleksAgata [21]

Answer:

6.39 J of energy is needed to generate 0.71 * 10⁻¹⁶ kg mass

Explanation:

According to the Equation: E = mc²

where the mass, m = 0.71 * 10⁻¹⁶ kg

the speed of light, c = 3 * 10⁸ m/s

The amount of energy needed to generate a mass of 0.71 * 10⁻¹⁶ kg is calculated as follows:

E = (0.71 * 10⁻¹⁶) (3 * 10⁸)²

E = 0.71 * 10⁻¹⁶ * 9 * 10¹⁶

E = 0.71 * 9

E = 6.39 J

6 0

3 years ago