A 4 kg and 6 kg bowling ball are dropped from the same height at the same time. The two balls strike the ground at the same time
2 answers:
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Answer:
The charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C
Explanation:
Here is the complete question
Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.
Solution
The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by
F = kq₁q₂/r₁²
q₁q₂ = Fr₁²/k
q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²
q₁q₂ = 6 × 10⁻¹⁶ C² (1)
When the charges are brought together, the charge is now q = (q₁ + q₂)/2
The new repulsive force F = 1.406 × 10⁻⁴ N at a distance of r₂ = 20 cm = 0.2 m is then
F₂ = kq²/r₂²
q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²
q² = 6.25 × 10⁻¹⁶ C²
q = √(6.25 × 10⁻¹⁶) C
q = 2.5 × 10⁻⁸ C
(q₁ + q₂)/2 = 2.5 × 10⁻⁸ C
(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C
q₁ + q₂ = 5 × 10⁻⁸ C (2)
q₁ = 5 × 10⁻⁸ C - q₂ (3)
Substituting equation (3) into (1), we have
(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²
Expanding the bracket, we have
(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²
So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0
Using the quadratic formula to find q₂
q₁ = 5 × 10⁻⁸ C - q₂
q₁ = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C
q₁ = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C
So the charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C
The acceleration of the wagon is found by applying Newton's Second Law of motion.
1. The responses for question 1 are;
- x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
- y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
- x-component of the tension in the rope on the left is -80.0 N
- y-component of the tension in the rope on the left is 0
2. The net force in the x-direction is approximately <u>11.93 N</u>
3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.
Reasons:
The given parameters are;
Mass of the wagon, m = 100.0 kg
Angle of inclination to the horizontal of the rope to the right, θ = 40.0°
Tension in the rope on the right = 120.0 N
Direction in which the rope on the left is pulled = To the west
Tension in the rope on the left = 80.0 N
1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;
x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>
y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>
The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;
x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>
y-component = 80.0 N × sin(180°) = <u>0.0 N</u>
2. The net force in the horizontal direction, Fₓ, is found as follows;
Fₓ = The x-component of the rope on the left + The x-component of the rope on the right
Which gives;
Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>
3. The net acceleration of the block is given as follows;
According to Newton's Second Law of motion, we have;
Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ
Fₓ = m × aₓ
Therefore;
- The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.
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