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Savatey [412]
2 years ago
6

Q1: An object with a charge of 1.2 C is located 4.5 m away from a second object that has a charge of 0.36 C. Find the electrical

force that the objects exert on each other. Use 8.988100 N·m2/C2 for the value of Coulomb's constant. Give your answer to two decimal places.
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
4 0

Answer:

a) F= 0,19  [N]   according to problem statement

b) F = 0,19*10⁹ [N]  using the right value of K

Explanation:

The force between two electric charges is according to Coulomb´s law is:

F = K * q₁*q₂ / d²    where  q₁  and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant  K = 8,988100*10⁹ N.m²/C². The problem establishes to use        K = 8,988100 N.m²/C².

NOTE: To value of is :  K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹

Then:

F = 8,988100 * 1,2* 0,36 / (4,5)²     [ N*m²/C² ] * [ C*C*/m²]

F = 3,882859/ 20,25  [N]

F= 0,19  [N]

The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies

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Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter

Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

R=\dfrac{1\times 10^{2}}{3.23}

R=30.95\ ohm

Now by Ohm's Law,

V= I\times R

Substituting the values we get

I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0
1 year ago
what is the resistance of a car light bulb that conducts 0.025A current when connected to a 12V car accumulator? is the current
Sophie [7]

One form of Ohm's Law says . . . . . Resistance = Voltage / Current .

R = V / I

R = (12 v) / (0.025 A)

R = (12 / 0.025) (V/I)

<em>R = 480 Ohms</em>

I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is.  (Floogle isn't sure either.)

If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.

If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.  

7 0
2 years ago
A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of
qaws [65]

Answer:

\frac{E_1}{E_2}= 4

Explanation:

Capacitance C is given by

C= \frac{\epsilon_0A}{d}

A= area of capacitor cross section

d= distance

therefore,

C_1= \frac{\epsilon_0A_1}{d_1}

A_1= πR^2

d_1= d

C_2= \frac{\epsilon_0A_2}{d_2}

A_= π(2R)^2

d_2 = 2d

q= \frac{\epsilon_0A_1}{d_1}V_1

threfore

V_1= \frac{qd_1}{\epsilon_0A_1}

and

V_2= \frac{qd_2}{\epsilon_0A_2}

also we know that E= V/d

⇒\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}

⇒\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}

= A_1/A_2= \frac{4R^2}{R^2}=4

therefore,

\frac{E_1}{E_2}= 4

5 0
2 years ago
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to
mote1985 [20]

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

6 0
1 year ago
An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far r from the center
Diano4ka-milaya [45]

Answer:

2023857702.507m

Explanation:

f=\frac{GMm}{r^{2} }

recall from newton's law of gravitation

G=gravitational constant

mshew=50g

melephant=5*10^3kg

rearth=radius of the earth 6400km or 6400000m

mearth= masss of the earth

Gm(shrew)m(earth)/r(earth)^2 = Gm(elephant)m(earth)/r^2

strike out the left hand side and right hand side variables

m(shrew)/r(earth)^2 = m(elephant)/r^2

r^2 = m(elephant).r(earth)^2 / m(shrew)       .........make r^2 the subject of the equation

r^2=(5*10^{3} *(6400000)^{2} )/.05

r^2=40960000000000

r=2023857702.507m

4 0
2 years ago
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