King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 12 m above the moat. T
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First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used are:
For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²
For constant velocity:
d = constant velocity*time
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v₁ - 0)/4.6 s
v₁ = 17.48 m/s
Total distance = d1 + d2 + d3
d1 = d = v,initial*t + 1/2*at²
d2 = constant velocity*time
Total distance = 0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
d3 = 56.69 m
For constant acceleration:
a = v,final - v,initial /t
d = v,initial*t + 1/2*at²
For constant velocity:
d = constant velocity*time
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v₁ - 0)/4.6 s
v₁ = 17.48 m/s
Total distance = d1 + d2 + d3
d1 = d = v,initial*t + 1/2*at²
d2 = constant velocity*time
Total distance = 0*(4.6) + 1/2*(3.8)(4.6)² + (17.48)(9.2) + d3= 257.71
d3 = 56.69 m
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