Question

coulombs law
formulas:
force (n) $f = k \frac{q_1 q_2}{r^2}$ charges (c)
constant (9 × 10⁹ n·m²/c²) distance (m)

1. a negative charge of - 5.0 x 10⁻⁴ c and a positive charge of 3.0 x 10⁻⁵ c are separated by 80m. what is the force between the two charges?

work and energy

formulas: work = force x distance

1. amy uses 100n of force to push a lawn mower 10 meters. how much work does she do?

1. angela uses a force of 25 newtons to lift her grocery bag while doing 10 joules of work. how far did she lift the grocery bags?

1. a bowling ball is sitting 4m above the floor. its mass is 6kg. what is its potential energy?

1. what is the same ball’s kinetic energy if it is moving at 4m/s?

Explanation:

Question 28:

Step1: Identify given values

\( q_1 = -5.0 \times 10^{-4} \, \text{C} \), \( q_2 = 3.0 \times 10^{-5} \, \text{C} \), \( r = 80 \, \text{m} \), \( K = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)

Step2: Apply Coulomb's Law formula

\( F = K \frac{|q_1 q_2|}{r^2} \) (we take absolute value for magnitude, sign indicates attraction/repulsion)
Substitute values:
\( F = 9 \times 10^9 \times \frac{|(-5.0 \times 10^{-4})(3.0 \times 10^{-5})|}{80^2} \)
First, calculate numerator: \( |(-5.0 \times 10^{-4})(3.0 \times 10^{-5})| = 1.5 \times 10^{-8} \)
Denominator: \( 80^2 = 6400 \)
Then, \( F = 9 \times 10^9 \times \frac{1.5 \times 10^{-8}}{6400} \)
\( F = 9 \times 10^9 \times 2.34375 \times 10^{-12} \)
\( F = 0.02109375 \, \text{N} \approx 0.021 \, \text{N} \) (attractive force since charges are opposite)

Step1: Identify formula and values

Formula: \( W = F \times d \)
\( F = 100 \, \text{N} \), \( d = 10 \, \text{m} \)

Step2: Substitute values

\( W = 100 \times 10 = 1000 \, \text{J} \)

Step1: Rearrange work formula

From \( W = F \times d \), solve for \( d \): \( d = \frac{W}{F} \)

Step2: Substitute values

\( W = 10 \, \text{J} \), \( F = 25 \, \text{N} \)
\( d = \frac{10}{25} = 0.4 \, \text{m} \)

Answer:

\( \approx 0.021 \, \text{N} \) (attractive)

Question 29: