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3 years ago

If 3.2*10^20 electrons pass through a wire in 4s, what would be the electrical current in the wire?

Physics

1 answer:

rewona [7]3 years ago

4 0

Answer: 12.8 A

Explanation:

Current is defined as the rate of flow of electric charge based on the formula:

I(current) = deltaQ(change in charge)/deltat(change in time).

First, however, we must convert the number of electrons into the number of coulombs. Based on the fact that the charge of 1 electron or 1 elementary charge is equal to 1.6*10^-19 C, we can calculate:

3.2*10^20 e = 1.6*10^-19*3*10^20 C = 51.2 C.

Now we use: I = Q/t = 51.2/4 = 12.8 A.

Hope this helped.

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an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s.where will the potential energy of the object b

Kay [80]

Answer:

333.3 m

Explanation:

Given

$m =100g\ =\ 0.1kg\\v = 100 m/s\\g = 10 m/s ^2$

Potential energy = $\frac{2}{3}\ of\ Kinetic\ energy$ ......Equation(1)

We know that

Potential energy=mgh

Kinetic energy = $\frac{1}{2} mv^{2}$

Now From the Equation(1)

$mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\ gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\ m$

3 0

2 years ago

An 11.0 -W energy-efficient fluorescent lightbulb is designed to produce the same illumination as a conventional 40.0-W incandes

yarga [219]

The amount or cost that the user of the energy-efficient bulb save during 100h of use will be $0.319.

<h3>How to calculate the cost?</h3>

For the 11.0W bulb, it should be noted that the value will be:

= 11.0 × 100 × (1/1000) × 0.110

= $0.121

The 40W bulb will be:

= 40 × 100 × (1/1000) × 0.110

= $0.44

Therefore, the amount that will be saved will be:

= $0.44 - $0.121

= $0.319

Learn more about cost on:

brainly.com/question/25109150

#SPJ4

6 0

2 years ago

A planet is discovered orbiting around a star in the galaxy Andromeda at four times the distance from the star as Earth is from

Dmitriy789 [7]

Answer:

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

Explanation:

The orbital period of a planet around a star can be expressed mathematically as;

T = 2π√(r^3)/(Gm)

Where;

r = radius of orbit

G = gravitational constant

m = mass of the star

Given;

Let R represent radius of earth orbit and r the radius of planet orbit,

Let M represent the mass of sun and m the mass of the star.

r = 4R

m = 16M

For earth;

Te = 2π√(R^3)/(GM)

For planet;

Tp = 2π√(r^3)/(Gm)

Substituting the given values;

Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)

Tp = 2π√(4R^3)/(GM)

Tp = 2 × 2π√(R^3)/(GM)

So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

5 0

2 years ago

Read 2 more answers

Ian walks 2 km to his best friends house, then walks 0.5 km to the library. He then makes 2.5 km walk home. The entire walk took

allochka39001 [22]

1. 2+0.5+2.5= 3. 2km/hr average

2. 14-6=4seconds. 8m/s in 4s = 2m/s acceleration

3. 15m/s divided by 2.5 = 6m/s acceleration

5 0

3 years ago

A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because

amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0

3 years ago