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3 years ago

What the unit for electric current

2 answers:

7 0

The unit of electric current is the Ampere.

1 Ampere of current means that if you set up your chair and
stare at the electrons flowing past one point in the circuit, you'll
see 1 coulomb of charge passing that point every second.

How will you recognize 1 coulomb of charge ?
Well, every electron carries the same amount of charge, and
we know how much that is. (Read about the Millikan oil-drop
experiment in 1909.) So we know how many electrons it takes
to carry 1 coulomb of charge past the point you're watching.

All you have to do is count the electrons as they zip past.
Every time you count 6,241,509,343,000,000,000 electrons,
you can tick off 1 coulomb of total charge that they're carrying.
If you reach that count every second, you know the current
passing that point is 1 Ampere.

horrorfan [7]3 years ago

3 0

Answer: amps

Explanation:

You might be interested in

A weather balloon is designed to expand to a maximum radius of 24 m at its working altitude, where the air pressure is 0.030 atm

nlexa [21]

Answer:

Radius at liftoff 8.98 m

Explanation:

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

First, we assume balloon is spherical in nature,

and that the working gas obeys the gas laws.

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere = $\frac{4}{3} \pi r^{3}$

volume of balloon = $\frac{4}{3}$ x 3.142 x $24^{3}$ = 57913.34 m^3

using the gas equation,

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.

imputing values, we have

$\frac{0.03*57913.34}{200}$ = $\frac{1*V2}{349}$

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = 3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.

from the formula volume of a sphere,

V = $\frac{4}{3} \pi r^{3}$ = $\frac{4}{3}$ x 3.142 x $r^{3}$ = 3031.76

4.19 $r^{3}$ = 3031.76

$r^{3}$ = 3031.76/4.19

radius r of the balloon on liftoff = $\sqrt[3]{723.57}$ = 8.98 m

4 0

3 years ago

A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I1 is the mo

Sati [7]

Answer:

The correct answer to the following question will be Option A (I1 > I2).

Explanation:

Method for moment of inertia because of it's viewpoint including object at a mean distance "r" from the axis is,

⇒ mr²

For Case 1:

Let the length of a rod be "r".

The axis passes via the middle of that same rod so that the range from either the axis within each dumbbell becomes " $\frac{r}{2}$ ".

Now,

Now total moment of inertia = sum of inertial moment due to all of the dumbbell

⇒ $l1=m(\frac{r}{2})^2+m(\frac{r}{2})^2$

⇒ $\frac{mr^2}{2}$

For Case 2:

Axis moves via one dumbbell because its range from either the axis becomes zero (0) and its impact is zero only at inertia as well as other dumbbell seems to be at a range "r" from either the axis

Now,

Total moment of inertia = moment of inertia of dumbbell at distance "r".

$l2=mr^2$

And now we can infer from this one,

⇒ $mr^2>\frac{mr^2}{2}$

So that "I1 > I2" is the right answer.

5 0

4 years ago

After your patched the roof, you dropped the hammer off the house and you heard it land on your toolbox on the ground 3.3s later

Schach [20]

Answer:

53.36 m.

Explanation:

Initial velocity of the toolbox u = 0

acceleration due to gravity g = 9.8 m s⁻²

time t = 3.3 s

height of roof h = ?

h = ut + 1/2 g t²

= 0 + .5 x 9.8 x 3.3²

= 53.36 m.

3 0

4 years ago

At a certain elevation, the ________ , the air becomes saturated and water-vapor molecules ________.

andriy [413]

At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.

As the altitude of air increase, the atmospheric pressure decrease due to which the temperature of the air decrease. The water molecules in the atmosphere start condensing, which saturate the air (that is air can no hold water molecules), due to which the water vapour molecules starts condensing and falls on the earth in the form of rain.

4 0

4 years ago

1) Find the voltage on a circuit with a resistance of 12.5 Ω if it has a current of 2.35 A.

KengaRu [80]

1. The voltage on the circuit with a resistance of 12.5 Ω and current of 2.35 A is 29.4 V.

2.The resistance in the circuit is found to be 1.45 Ω.

3. The equivalent resistance of the resistors connected in series is 24 Ω.

4. The equivalent resistance of the resistors connected in parallel is 2.18Ω.

5. The power generated is 552.5 W.

6. The frequency of the green light is 0.56×10¹⁵ Hz.

Explanation:

1) This problem can be solved using Ohm's law. Here the resistance (R) of the circuit is given to be 12.5 Ω and the current (I) is stated to be 2.35 A. So the ohm's law states that in a closed circuit, the voltage will be directly proportional to the current flowing in the circuit and the resistance will act as the proportionality constant.

$V = I * R = 2.35*12.5 = 29.4 V$

So, the voltage on the circuit with a resistance of 12.5 Ω and current of 2.35 A is 29.4 V.

2) Using the same Ohms' law, now we have to determine the resistance. So in this case, the voltage is given as 9 V and the current is said to be 6.2 A, then resistance can be determined as the ratio of voltage to current.

$R = \frac{V}{I} =\frac{9}{6.2} =1.45 Ohms$

So, the resistance in the circuit is found to be 1.45 Ω.

3) Here, the resistances of three resistors are given as 4 Ω, 8 Ω and 12 Ω. And it is stated that the resistances are connected or wired in series. Then the equivalent resistance will be obtained by the sum of resistances of three resistors, as the current flow will be constant in all the three resistors.

$R_{s} = R_{1} + R_{2} + R_{3} \\ \\R_{s} = 4+8+12 = 24 ohms$

Thus, the equivalent resistance of the resistors connected in series is 24 Ω.

4) Now, if the resistors are connected in parallel, then the equivalent resistance will be ratio of product of resistances to the sum of the resistances.

$\frac{1}{R_{p} }= \frac{1}{R_{1} } + \frac{1}{R_{2} } +\frac{1}{R_{3} }\\\\\frac{1}{R_{p} }=\frac{1}{4}+ \frac{1}{8} +\frac{1}{12} = \frac{6+3+2}{24} =\frac{11}{24} \\\\R_{p} = \frac{24}{11 } =2.18 Ohm$

Thus, the equivalent resistance of the resistors connected in parallel is 2.18Ω.

5) Power generated by the person can be obtained by the ratio of work done by the person to the time in which the work is done. So the work done can be obtained by the product of force with displacement.

As here the weight lifted by the person will act as dominant force on the person. So the force is considered as F = 956 N and the displacement is d = 2.41 m, then

$Work done = Force * displacement = 956*2.41 =2303.96 J$

So, the work done is obtained as 2303.96 J and the time is given as 4.17 s, then

$Power = \frac{Work done}{Time} =\frac{2303.96}{4.17} =552.5 W$

So, the power generated is 552.5 W.

6) In this, the wavelength of green light is given as 5.34 × 10⁻⁷ m. It is known that the wavelength is inversely proportional to the frequency.

$Wavelength = \frac{Speed of light}{Frequency}$

As, speed of light is known as 3×10⁸ m/s, the frequency will be determined as the ratio of speed of light to wavelength.

$Frequency = \frac{Speed of light}{Wavelength} =\frac{3*10^{8} }{5.34*10^{-7} } \\\\Frequency =0.56*10^{15} Hz$

Thus, the frequency of the green light is 0.56×10¹⁵ Hz.

8 0

4 years ago