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Solnce55 [7]
3 years ago
7

On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of

electrons? (b)What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?
Physics
1 answer:
Advocard [28]3 years ago
5 0

Answer:

On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (<em>the mass will decrease by a very small factor</em>)

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?  (<em>The mass will increase by a very small factor</em>)

Explanation:

(a) On a tiny scale, what happens to an initially neutral object’s mass when it gains a net positive charge through the exchange of electrons? (<em>the mass will decrease by a very small factor</em>)

The mass of an atom is given by the sum of the masses of the protons, neutrons and electrons. Electrons has lower mass than protons and neutrons, so they have a minor contribution to the total mass of the atom.    

When an object is electrically neutral it means that it has the same number of protons and electrons. For the case of an object positively charged, the rate of protons is greater than the number of electrons. That means that atom lose electrons so the mass will decrease in a very small factor.

(b) What happens to the mass of an initially neutral object when it gains a net negative charge through the exchange of electrons?  (<em>The mass will increase by a very small factor</em>)

For the case when the object is negatively charged, it means that the atom gains electrons from another object, leading to the conclusion that the mass of the atom will increase in a very small factor.  

Key values:

Electron mass: 9.1095×10⁻³¹ Kg

Proton mass: 1.67261×10⁻²⁷ Kg

Neutron mass: 1.67492×10⁻²⁷ Kg

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STatiana [176]

Answer:

<h3>The answer is 30.43 L</h3>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{35000 \times 80}{92000}  =  \frac{2800000}{92000}  =   \frac{700}{23}  \\  = 30.434782...

We have the final answer as

<h3>30.43 L</h3>

Hope this helps you

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A lithium atom has three protons, three neutrons, and three electrons. What is the overall charge on this atom?
Marizza181 [45]

Answer:

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Explanation:

The overall charge on this atom is 0.

  To find the charge on an atom;

           charge  = number of protons - number of electrons.

Note:

  • Protons are the positively charged particles in an atom
  • Electrons are the negatively charged particles in an atom
  • Neutrons carries no charges on them.

Since the atom is made up of equal number of protons and electrons, the charge on it is 0.

If the number of electrons is more, the atom will be negatively charge but if the number of protons is more, it will be positively charged.

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what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?
zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

R = ρL/A ......... (1)

A = πr² (since the wire is circular in shape)

r = d/2

A = πr² = π(d/2)²

A = πd²/4

Substitute the value of A into equation 1

R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

R = 4ρL /πd²

Where:

R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

I is the current

From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

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