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(1) A positive charge +3 C is separated from another positive charge of +5 C by a distance of 7m. What is the magnitude of the e

Aneli [31]

1. The magnitude of the electric force between the two charges is 2.8×10⁹ N (Option B)

2. The net charge on the molecule is -8×10⁻¹⁹ C (Option D)

3. The magnitude of the force between the charges is 16000 N (Option C)

4. The correct statement is: A neutral object has equal numbers of protons and electrons. (Option C)

<h3>1. How to determine the force</h3>

Charge 1 (q₁) = +3 C
Charge 2 (q₂) = +5 C
Electric constant (K) = 9×10⁹ Nm²/C²
Distance apart (r) = 7 m
Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 3 × 5) / (7)²

F = 2.8×10⁹ N

<h3>2. How to determine the net charge on the molecule</h3>

Electron = 223 electrons
Proton = 218 protons
Net Charge =?

Charge = Proton - Electron

Charge = 218 - 223

Charge = -5 electrons

But

1 electron = 1.6×10⁻¹⁹ C

Thus,

Net Charge = -5 × 1.6×10⁻¹⁹ C

Net Charge = -8×10⁻¹⁹ C

<h3>3. How to determine the force</h3>

Charge 1 (q₁) = 2×10⁻⁴ C
Charge 2 (q₂) = 8×10⁻⁴ C
Electric constant (K) = 9×10⁹ Nm²/C²
Distance apart (r) = 0.3 m
Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 2×10⁻⁴ × 8×10⁻⁴) / (0.3)²

F = 16000 N

<h3>4. What is a neutral object?</h3>

A neutral object is an object having equal numbers of protons and electrons. For example, an object with 4 protons and 4 electrons is said to be neutral as illustrated below

Electron = 4 electrons
Proton = 4 protons
Net Charge =?

Charge = Proton - Electron

Charge = 4 - 4

Charge = 0 (neutral)

Thus, the correct statement about neutral object, given in the question is: A neutral object has equal numbers of protons and electrons (Option C)

Learn more about Coulomb's law:

brainly.com/question/506926

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6 0

2 years ago

A 67 kg kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m m . What is the effective s

amid [387]

Point of correction, spring constant is 2.3×10−2 m not 2.3×10−2 m m

Answer:

28577 N/m

Explanation:

From Hooke's law, F=kx where F is force applied, k is spring constant and x is compression

F=mg=67*9.81

$k=\frac {mg}{x}$

$k=\frac {67*9.81}{2.3\times 10^{-2}}=28576.95652 N/m$

Approximately, 28577 N/m

8 0

3 years ago

A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a

blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

$E=7*10^{9}N/C$

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field is mathematically given as

$E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\$

$E=7*10^{9}N/C$

For more information on this visit

brainly.com/question/21811998

4 0

3 years ago

Elle is playing with a ball in a bus that moves in a straight line with constant velocity. What can you say about the motion of

kramer

Given that they are all on the same bus that is travelling in a straight line at the same velocity, when Elle throws the ball directly upwards, the ball will simply fall back to her. This is because the bus, Elle, and the ball are all travelling in the same direction and at the same speed. Among the choices, the correct answer is A.

8 0

3 years ago

1.A car starts from rest and acquires a velocity of 54 km/h in 2 minutes.Find(i) the acceleration and(ii) distance travelled by

Sergeu [11.5K]

Answer:

1.) 1620 km/h^2

2.) 2.7 km

Explanation:

1.) Given that the car start from rest. The initial velocity U will be equal to zero. That is,

U = 0.

Final velocity V = 54 km/h

Time t = 2 minute = 2/60 = 1/30 hour

Acceleration a will be change in velocity per time taken. That is,

a = ( V - U )/ t

Substitute V, U and t into the formula

a = 54 ÷ 1/30

a = 54 × 30 = 1620 km/h ^2

2.) Distance travelled S by the car during the time can be calculated by using the 2nd equation of motion.

S = Ut + 1/2at^2

Substitute all the parameters into the formula

S = 54 × 1/30 + 1/2 × 1620 × (1/30)^2

S = 54/30 + 810 × 1/900

S = 54/30 + 810/900

S = (1620+810)/900

S = 2430/900

S = 2.7 km.

Therefore, distance travelled by the car during this time is 2.7 km

4 0

3 years ago