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hram777 [196]
3 years ago
13

Compare the numbers of protons and electrons in a positive ion with the numbers of protons and electrons in a negative ion.

Physics
1 answer:
kow [346]3 years ago
6 0

Answer:

In a positive ion, the number of protons is larger than the number of electrons.

In a negative ion, the number of protons is smaller than the number of electrons.

Explanation:

Each proton carries a positive charge of one unit.

Each elec in tron carries a negative charge of one unit.

In an atom, there are as many protons as electrons. Hence, they are neutral.

However, in a positive ion, there are less negative charge than positive charge. Hence the net charge is positive. That also means that there are fewer negatively-charged electrons than positively-charged protons.

Similarly, in a negative ion, there are more negative charge than positive charge. Hence the net charge is negative. That also means that there are more negatively-charged electrons than positively-charged protons.

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A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
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Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

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An object is moving at a constant velocity of 4 m/s. What is it's initial velocity?
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Wood burning in a fire is what property
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4 0
3 years ago
Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. F
Art [367]

Answer:

37.8 m

Explanation:

At point 0, the ball is at height y₀.

At point 1, the ball is at height 30 m.

At point 2, the ball is at height 0 m.

Given:

y₁ = 30 m

y₂ = 0 m

v₀ = 0 m/s

a = -10 m/s²

t₂ − t₁ = 1.5 s

Find: y₀

Use constant acceleration equation.

y = y₀ + v₀ t + ½ at²

Evaluate at point 1.

y₁ = y₀ + v₀ t₁ + ½ at₁²

30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²

30 = y₀ − 5t₁²

Evaluate at point 2.

y₂ = y₀ + v₀ t₂ + ½ at₂²

0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²

0 = y₀ − 5t₂²

y₀ = 5t₂²

Substitute:

y₀ = 5 (1.5 + t₁)²

y₀ = 5 (2.25 + 3t₁ + t₁²)

y₀ = 11.25 + 15t₁ + 5t₁²

30 = 11.25 + 15t₁ + 5t₁² − 5t₁²

30 = 11.25 + 15t₁

t₁ = 1.25

30 = y₀ − 5t₁²

30 = y₀ − 5(1.25)²

y₀ ≈ 37.8

4 0
3 years ago
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