All categories

3 years ago

Why does current flow through a circuit?

2 answers:

6 0

WHAT MAKES CURRENT FLOW IN A CIRCUIT? When a wire is connected to battery terminals, electrons flow from negative to positive. Unlike (opposite) charges attract, like (same) charges repel. Electrons have a negative charge—they are repelled from the negative and attracted to the positive.

Ghella [55]3 years ago

3 0

Answer:

Explanation:

cause it do

You might be interested in

A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

$\Phi=B\ A\ cos\theta$

We are told the values of $\Phi$ and B, then we can calculate the magnitude

$\frac{\Phi}{A}=B\ cos\theta$

3 0

3 years ago

How long would this acceleration last

Aleksandr-060686 [28]

It would last as long as the applied force continued, or until the accelerating object hit something.

6 0

3 years ago

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago

What are the units of radiation?

Ksenya-84 [330]

The radiation dose absorbed by a person is measured using the conventional unit rad or the SI unit gray. The biological risk of exposure to radiation is measured using the conventional unit rem or the SI unit sievert.

I hope this helps!

3 0

3 years ago

A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease

skelet666 [1.2K]

The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

$P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm$

So, the new pressure is 2 atm.

7 0

3 years ago