A student has an object with a charge of -3 Coulombs. Which of the following objects would have the strongest attraction to object A?

Question

options:

An object with a -3 charge at a distance of 5 cm
An object with a -3 charge at a distance of 5 cm

An object with a +3 charge at a distance of 5 cm
An object with a +3 charge at a distance of 5 cm

An object with a +3 charge at a distance of 2 cm
An object with a +3 charge at a distance of 2 cm

An object with a -4 charge at a distance of 2 cm

Answer 1

To determine which object will have the strongest attraction to the object with a charge of -3 Coulombs, we will consider both the magnitudes of the charges and the distances between the objects.

The force of attraction (or repulsion) between two charges can be determined using Coulomb's Law:

\[ F = k \frac{|q_1 q_2|}{r^2} \]

Where:

• \( F \) is the force between the charges,
• \( k \) is Coulomb's constant,
• \( q_1 \) and \( q_2 \) are the magnitudes of the two charges,
• \( r \) is the distance between the centers of the two charges.

Let's analyze the options:

1. An object with a -3 charge at a distance of 5 cm:

   • Charge: \(-3 C\), Distance: \(5 cm\)
   • \( F = k \frac{|(-3)(-3)|}{(0.05)^2} = k \frac{9}{0.0025} \) (repulsive, so we will not consider it for attraction)

2. An object with a -3 charge at a distance of 5 cm (repeat option, same reasoning applies):

   • Same calculation as above; it will also repel, not attract.

3. An object with a +3 charge at a distance of 5 cm:

   • Charge: \(+3 C\), Distance: \(5 cm\)
   • \( F = k \frac{|(-3)(3)|}{(0.05)^2} = k \frac{9}{0.0025} = k \times 3600 \)

4. An object with a +3 charge at a distance of 2 cm:

   • Charge: \(+3 C\), Distance: \(2 cm\)
   • \( F = k \frac{|(-3)(3)|}{(0.02)^2} = k \frac{9}{0.0004} = k \times 22500 \)

5. An object with a -4 charge at a distance of 2 cm:

   • Charge: \(-4 C\), Distance: \(2 cm\)
   • \( F = k \frac{|(-3)(-4)|}{(0.02)^2} = k \frac{12}{0.0004} = k \times 30000 \) (this is also a repulsive interaction)

Based on the calculations:

• The attraction to \( +3 C \) at \( 2 cm \) gives \( k \times 22500 \).
• The attraction to \( +3 C \) at \( 5 cm \) gives \( k \times 3600 \).
• The -4 C charge at \( 2 cm \) gives a repulsive force of \( k \times 30000 \) which we will not consider.

Therefore, the object with the strongest attraction to object A is:

An object with a +3 charge at a distance of 2 cm, as it provides the largest attractive force \( k \times 22500 \) when taking both the charge and distance into account.

Answer 2

On Earth, when you pick up a bowling ball from the ground it feels very heavy and takes a lot of force to lift it. However on the moon the bowling ball feels lighter and easier to pick up. Why does a bowling ball feel easier to pick up on the moon than on Earth

options:

The weight of the bowling ball is less on the moon than on Earth
The weight of the bowling ball is less on the moon than on Earth

The mass of the bowling ball is greater on the moon than on Earth
The mass of the bowling ball is greater on the moon than on Earth

The mass of the bowling ball is less on the moon than on Earth
The mass of the bowling ball is less on the moon than on Earth

The weight of the bowling ball is greater on the moon than on Earth

Answer 3

To understand why a bowling ball feels lighter on the Moon than on Earth, we should clarify the concepts of weight and mass.

• Mass is a measure of the amount of matter in an object and is constant regardless of where the object is located.
• Weight is the force exerted on an object due to gravity and is affected by the local gravitational field strength. The formula for weight is:

\[ \text{Weight} = \text{mass} \times \text{gravitational acceleration} \]

Gravitational Acceleration

• On Earth, the gravitational acceleration is approximately \(9.81 , \text{m/s}^2\).
• On the Moon, the gravitational acceleration is about \(1.62 , \text{m/s}^2\), which is roughly \(1/6\)th of that on Earth.

Conclusion

Since the mass of the bowling ball does not change regardless of location, its mass remains the same whether it is on Earth or the Moon. However, the weight (the force due to gravity) is less on the Moon because the gravitational pull is weaker.

Therefore, the correct statement is:

The weight of the bowling ball is less on the moon than on Earth.

Thus, it feels easier to pick up the bowling ball on the Moon due to its reduced weight.

Answer 4

If the mass of a planet is cut in half, how would the force of gravity between that planet and the Sun change?

options:

The force of gravity would increase four times
The force of gravity would increase four times

The force of gravity would not change
The force of gravity would not change

The force of gravity would double
The force of gravity would double

The force of gravity would decrease

Answer 5

To understand how the force of gravity between a planet and the Sun changes if the mass of the planet is cut in half, we can use Newton's Law of Universal Gravitation, which is given by the formula:

\[ F = G \frac{m_1 m_2}{r^2} \]

Where:

• \( F \) is the gravitational force between two masses,
• \( G \) is the gravitational constant,
• \( m_1 \) and \( m_2 \) are the masses of the two objects (in this case, the mass of the planet and the mass of the Sun), and
• \( r \) is the distance between the centers of the two masses.

In this scenario, if we cut the mass of the planet (let's call it \( m_p \)) in half, the new mass of the planet becomes \( \frac{1}{2} m_p \). The mass of the Sun (\( m_s \)) and the distance (\( r \)) remain unchanged.

So the new gravitational force \( F' \) can be expressed as:

\[ F' = G \frac{\left(\frac{1}{2} m_p\right) m_s}{r^2} = \frac{1}{2} \left(G \frac{m_p m_s}{r^2}\right) = \frac{1}{2} F \]

This shows that the gravitational force \( F' \) between the planet and the Sun would be half of the original force \( F \).

Conclusion

Thus, if the mass of the planet is cut in half, the force of gravity between that planet and the Sun would decrease.

The correct answer is:

The force of gravity would decrease.