<span>The distance between two objects is increased by three times the oringinal distance. Since they were already separated by one time the original distance,
the additional three times the oringinal distance now puts them four times the original distance apart.
Whether we're talking about the gravitational forces of attraction or
the electrical forces of attraction, either one is inversely proportional
to the square of the distance between the objects.
So changing the distance to four times the original distance causes
the forces to become 1/4</span>² as strong as they were originally.
The forces become 1/16 of their original magnitude.<span>
</span>
Answer:
Option (c) : 20°C
Explanation:
T(final) = 500* 10 + 100*70/600 = 20°C
Average Velocity = Total Displacement / Total time
1st part of journey, 350 km at velocity 125 km/h
Time = 350 / 125 = 2.8 hours.
2nd part of journey, 220 km at velocity 115 km/h
Time = 220 / 115 = 1.9 hours
Average Velocity = Total Displacement / Total time
= (350 + 220) / (2.8 + 1.9)
= 570 / 4.7 ≈ 121.3 km/hr
Average Velocity ≈ 121 km/hr due south.
Option C.
Answer:
a) a = - 0.106 m/s^2 (←)
b) T = 12215.1064 N
Explanation:
If
F₁ = 9*1350 N = 12150 N (→)
F₂ = 9*1365 N = 12285 N (←)
∑Fx = M*a = (M₁ +M₂)*a (→)
F₁ - F₂ = (M₁ +M₂)*a
→ a = (F₁ - F₂) / (M₁ +M₂ ) = (12150-12285)N/(9*68+9*73)Kg
→ a = - 0.106 m/s^2 (←)
(b) What is the tension in the section of rope between the teams?
If we apply ∑Fx = M*a for the team 1
F₁ - T = - M₁*a ⇒ T = F₁ + M₁*a
⇒ T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
If we choose the team 2 we get
- F₂ + T = - M₂*a ⇒ T = F₂ - M₂*a
⇒ T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
