Sum of number of protons and neutrons in the nucleus

What happens to the strength of the magnetic field as you come closer to the current carrying wire?

lina2011 [118]

Electric energy is carried by current, which is the flow of electrons, which are the negatively charged sub-particles of atoms. This transfer of electrons from one place to another powers our lights, phones, appliances and many other things we use every day. Another interesting phenomenon of flowing current is that it produces its own magnetic field. Electricity and magnetism are very closely linked in that all closed loop currents create their own magnetic fields, and magnetic fields acting upon closed loop circuits can produce current or even change current diction.

8 0

2 years ago

You walk 200m south, then 50m east. What is your displacement?

vredina [299]

250m good luck onnur test

5 0

3 years ago

Is the following equation balanced?<br><br> 2Al + 3NiBr2 AlBr3 + 3Ni<br><br> yes<br> no

jeka94

<h3><u>Answer;</u></h3>

<u>No-</u>the equation is not balanced.

<h3><u>Explanation;</u></h3>

The balanced equation will be;

<em><u>According to the law of conservation of mass, the mass of reactants should always be the same as the mass of the products in a chemical equation. </u></em>
Therefore, the number of atoms of each element in a chemical equation should always be the same on both sides of the equation, that is the side of reactants and side of products.
<em><u>Balancing of chemical equations ensures that the number of atoms of each element is equal in both sides of the equation</u></em>.

5 0

3 years ago

Read 2 more answers

The mass of the earth is 5.96/10kg^24. The radius of the earth is approximately 6.37x10^6 calculate the force of gravity

n200080 [17]

<h2>Answer:g=9.79 $ms^{-2}$ ,A object of mass $m$ at the surface of earth experiences a force $mg$ </h2>

Explanation:

Let $M$ be the mass of earth.

Let $R$ be the radius of earth.

Let $G$ be the universal gravitational constant.

Given,

$M=5.96\times 10^{24}Kg$

$R=6.37\times 10^{6}m$

$G=6.67259 \times 10^{-11}$ $Nm^{2}Kg^{-2}$

Let $g$ be the acceleration due to gravity.

Then, $g=\dfrac{GM}{R^{2}}$

$g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}$

$g=9.79ms^{-2}$

A object of mass $m$ at the surface of earth experiences a force $mg$

3 0

3 years ago

An eagle flying at 15 m/s has 600 J of kinetic energy. About how much is the eagles mass?

Eduardwww [97]

Answer:

5.33kg

Explanation:

Given parameters:

Velocity of eagle = 15m/s

Kinetic energy of the eagle = 600J

Unknown:

Mass of the eagle = ?

Solution:

The kinetic energy of any body is the energy due to the motion of a body. There are different forms of kinetic energy some of which are thermal, mechanical, electrical energy.

The formula of kinetic energy is given as;

Kinetic energy = $\frac{1}{2}$ m v²

where m is the mass, V is the velocity

substitute the parameters in the equation;

600 = $\frac{1}{2}$ x m x 15²

225m = 1200

m = $\frac{1200}{225}$ = 5.33kg

3 0

2 years ago