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3 years ago

How much resistance is required to limit the current from a 12 V battery to 3.6 mA?

Physics

1 answer:

hammer [34]3 years ago

5 0

Answer:

3333.3 ohm

Explanation:

V=IR so R =V/I (in Ampere) so convert 3.6 mA to 3.6*10^-3 A

so R =12/3.6*10^-3 =3333.3

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Oduvanchick [21]

Answer

D. Base units

Explanation

Basic units are also called fundamental units. They are the standard units agreed internationally for measurements. Most of these measurements are taken from the ground and they are used to derive other units. They are seven in number. There are:

The metre (m)

The kilogram (kg)

The second (s)

The ampere (A)

The kelvin (K)

The candela (cd)

The mole (mol)

6 0

3 years ago

The composition of the universe only consists of 596 of ordinary matter. True O False

brilliants [131]

True
Because I know lol they make u type so much

6 0

3 years ago

Read 2 more answers

You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You disc

FrozenT [24]

Answer:

The balloon hit the ground with velocity -15.34 m/s

Explanation:

Lets explain how to solve the problem

You found that the best height to pitch a water balloon in order for it to

burst when it hits the ground is 12 meters.

We consider that the 12 meters is the maximum height, so the velocity

at this height is zero.

To find the velocity when the balloon hits the ground lets use the rule

v² = u² + 2gh, where v is the final velocity, u is the initial velocity, g is

the acceleration of gravity and h is the height.

u = 0 , h = 12 m , g = 9.8 m/s²

Substitute these values in the equation above

v² = 0 + 2(9.8)(12)

v² = 235.2

Take square root for both sides

v = ± $\sqrt{235.2}$

The velocity is downward, then it's a negative value

Then v = -15.34 m/s

The balloon hit the ground with velocity -15.34 m/s

6 0

3 years ago

For the PE formula, why is the height required for calculations? Why do we need to know the height in order to determine PE? *

Fudgin [204]

Answer:

Answer in Explanation

Explanation:

Whenever we talk about the gravitational potential energy, it means the energy stored in a body due to its position in the gravitational field. Now, we know that in the gravitational field the work is only done when the body moves vertically. If the body moves horizontally on the same surface in the Earth's Gravitational Field, then the work done on the body is considered to be zero. Hence, the work done or the energy stored in the object while in the gravitational field is only possible if it moves vertically. This vertical distance is referred to as height. This is the main reason why we require height in the P.E formula and calculations.

The derivation of this formula is as follows:

Work = Force * Displacement

For gravitational potential energy:

Work = P.E

Force = Weight = mg

Displacement = Vertical Displacement = Height = h

Therefore,

P.E = mgh

5 0

3 years ago

To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance

mart [117]

Answer:

a. 120 W

b. 28.8 N

Explanation:

To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance. Suppose a cyclist is traveling at 15 km/h on level ground. Assume he is using 480 W of metabolic power.

a. Estimate the amount of power he uses for forward motion.

b. How much force must he exert to overcome the force of air resistance?

(a) He is 25% efficient, therefore the cyclist will be expending 25% of his power to drive the bicycle forward

Power = efficiency X metabolic power

= 0.25 X 480

= 120 W

(b)

power if force times the velocity

P = Fv

convert 15 km/h to m/s

v = 15 kmph = 4.166 m/s

F = P/v

= 120/4.166

= 28.8 N

definition of terms

power is the rate at which work is done

force is that which changes a body's state of rest or uniform motion in a straight line

velocity is the change in displacement per unit time.

3 0

3 years ago