_________ on a body is produced by the gravitational force of attraction
1 answer:
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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 value of tension on the cable T = 1065.6 N
Explanation:
Mass = 888 kg
Initial velocity ( u )= 0.8
Final velocity ( V ) = 0
Distance traveled before come to rest = 0.2667 m
Now use third law of motion =
- 2 a s
Put all the values in above formula we get,
⇒ 0 = - 2 × a ×0.2667
⇒ a = 1.2
This is the deceleration of the box.
Tension in the cable is given by T = F = m × a
Put all the values in above formula we get,
T = 888 × 1.2
T = 1065.6 N
This is the value of tension on the cable.
5.5 s
Explanation:
The time it takes for the ball to reach its maximum height can be calculated using
since at the top of its trajectory. Plugging in the numbers,