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1 year ago

An LDR was placed near a light source. The following results were

Physics

1 answer:

Zina [86]1 year ago

6 0

Answer:

V = 440I

Explanation:

According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Let V be the potential difference

I is the current

R is the resistance

Mathematically;

V∝I

V = IR

Given

Potential difference V = 5.50Volts

Current = 12.5mA = 12.5*10^-3A

Current I = 0.0125A

Recall that V = IR

R = V/I

Substitute the given parameters into the formula;

R = 5.50/0.0125

R = 440 ohms

Hence the equation that links current, potential difference and resistance is expressed as;

V = I(440)

V = 440I

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How is Coulomb’s law similar to newton’s law of gravitational force? How is it different

natulia [17]

The similarities and the differences between gravitational and electric force are listed below

Explanation:

- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

By comparing the two equations, we find the following similarities:

Both the forces are inversely proportional to the square of the distance between the two objects, $F\propto \frac{1}{r^2}$
Both the forces are proportional to the product between the "main quantity" of each force, which is the mass for the gravitational force ( $F\propto m_1 m_2$ ) and the charge for the electric force ( $F\propto q_1 q_2$

Instead, we have the following differences:

The gravitational force is always attractive, since the sign of $m$ is always positive, while the electric force can be either attractive or repulsive, since the sign of $q$ can be either positive or negative
The value of the gravitational costant G is much smaller than the value of the Coulomb's constant, so the gravitational force is much weaker than the electric force

Learn more about gravitational force and electric force:

brainly.com/question/1724648

brainly.com/question/12785992

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

2 years ago

Which best compares a brush and a commutator in an electric motor?

alisha [4.7K]

The purpose of the brush is supposed to be to lessen the amount of damage or wear placed upon the motor. I hope this helps with your answer choices since you did not post any.

3 0

2 years ago

Read 2 more answers

A hydrogen atom that has lost its electron is moving east in aregion where the magnetic field is directed from south to north. I

Kitty [74]

Answer:

a. up

Explanation:

As per the rule of Fleming left hand, the three fingers should be places in a perpendicular manner i.e. mutually also.

The fore finger depicts the field direction

The middle finger depicts the velocity

And, the thumb finger depicts the force direction that experienced on that particle i.e. charged

So the electrons would be deflects to up

Hence, the correct option is a.

5 0

2 years ago

Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the

Alenkasestr [34]

Answer:

The acceleration of car 2 is four times of the acceleration of car 1.

Explanation:

The centripetal acceleration of the object is possessed when it moves in a circular path. It is given by :

$a=\dfrac{v^2}{r}$

In this case, two race cars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the track, but car 2 is driving twice as fast as car 1.

So,

$\dfrac{a_1}{a_2}=\dfrac{v_1^2}{v_2^2}$