A rock is thrown horizontally off a cliff with an initial speed of 3 m/s. The initial position of the rock is 10 meters above th
1 answer:
You might be interested in
Answer:
D = 271.54 m
Explanation:
given,
1. car accelerates at 4.6 m/s² for 6.2 s
2. constant speed for 2.1 s
3. slows down at 3.3 m/s²
distance travel for case 1
using equation of motion
d₁ = 88.41 m
case 2
constant speed for 2.1 s now, we have to find velocity
v = u + at
v = 0 + 4.6 x 6.2
v = 28.52 m/s
distance travel in case 2
d₂ = v x t
d₂ = 28.52 x 2.1 = 59.89 m
for case 3
distance travel by the car
v² = u² + 2 a s
final velocity if the car is zero
0² = 28.52² + 2 x (-3.3) x d₃
6.6 d₃ = 813.39
d₃ = 123.24 m
total distance travel by the car
D = d₁ + d₂ + d₃
D = 88.41 + 59.89 + 123.24
D = 271.54 m
Answer:
Average speed = 0.0075 m/s
Average velocity = 0.0025 m/s along forward direction
Explanation:
Speed is the ratio of distance and time and velocity is the ratio of displacement and time.
Distance traveled = 10 + 5 = 15 cm = 0.15 m
Displacement = 10 - 5 = 5 cm = 0.05 m
Time = 20 seconds
Average speed = 0.0075 m/s
Average velocity = 0.0025 m/s along forward direction
Answer:
Spring constant, k = 283.33 N/m
Explanation:
Given that,
Force acting on the spring, F = 8.5 N
Stretching in the spring, x = 3 cm = 0.03 m
Let k is the spring constant of the spring. It can be calculated using Hooke's law as :
k = 283.33 N/m
So, the spring constant of the spring is 283.33 N/m. Hence, this is the required solution.