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vodomira [7]
3 years ago
7

What affect would using a 12V car battery have on the operation of your circuit? (Do not try this.) What would happen to the cur

rent? What would happen to the resistance? Be specific and include appropriate calculations to support your answers.
Physics
1 answer:
k0ka [10]3 years ago
6 0

Answer:

Incomplete question

This is the completed question

If the resistor in the circuit had a larger resistance then the current would be then have to be proportionally smaller. Because the batteries each give off 1.5 volts then the current would have to be the variable that would change. What affect would using a 12V car battery have on the operation of your circuit? (Do not try this.) What would happen to the current? What would happen to the resistor?

Explanation:

Using ohms law as our basis

Ohms law state that, the voltage in an ohmic conductor is directly proportional to the current

V∝I

Resistance is the constant of proportionality

Then

V=iR

Since we want a relationship between current and resistance.

then, I=V/R

So, current is inversely proportional to Resistance

as the current increase the resistance reduce and as the current reduces the resistance increases.

a. So, increasing the voltage from 1.5V to 12V increases the current In the circuit because voltage Is directly proportional to I.

From ohms law

V=iR

When v =1.5V

I=1.5/R

When V increase to 12V

I=12/R

I.e, it increases by a factor of 8. Eight times it's initial value

b. Now, the resistance in the circuit is the constant of proportionality and it doesn't change in a given circuit expect when using a variable resistoa r like rheostat.

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Which of the following is NOT true regarding the far side of the moon:
frutty [35]
It contains no large maria
8 0
4 years ago
what is the magnitude of the vector described below? 13 m/s to the east a. east b. meters per second c. 13 m/s d. meters
Katen [24]

Answer:

C

Explanation:

Magnitude of any quantity is the measurable value of the quantity. While the direction of the given quantity is the specific pointing direction of position or the angle at which it move.

The magnitude of the vector described below? 13 m/s to the east will be 13 m/s

While the direction will be eastward.

Therefore, the magnitude is 13 m/s

The correct answer is option C

3 0
3 years ago
A weight lifter does 700J of work on a weight that he lifts in 3.1s. What is the power with
IgorLugansk [536]

Power = Work done/Time taken

So, keeping this in mind,we can solve it as follows:

= 700/3.1

= 7000/31

= 225.80 W

8 0
3 years ago
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov
son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

Where f_e is the eyepiece focal length of the microscope

  Now  making f_e the subject  of the formula

         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

         f_e  = 0.027 \ m

     

5 0
3 years ago
A 50.0 kg object is moving at 18.2 m/s when a 200 N force
gayaneshka [121]

Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity  = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a  = force ÷ mass

a = \frac{-200}{50}   =  -4 m/s²

we get here now required time that is

required time = \frac{V_{(final)} - V_{(initial)}}{a}     ...............1

put here value

required time = \frac{12.6-18.2}{-4}  

so distance will be

distance = \frac{V_{(final)}^2 - V_{(initial)}^2}{2a}    ........2

distance = \frac{12.6}^2 -{18.2}^2}{2\times (-4)}  

distance = 21.56 m

7 0
3 years ago
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