Answer:
1317.4 m
Explanation:
We are given that
Angle=
Initial speed =
We have to find the horizontal distance covered by the shell after 5.03 s.
Horizontal component of initial speed=
Vertical component of initial speed=
Time=t=5.03 s
Horizontal distance =
Using the formula
Horizontal distance=
Horizontal distance=1317.4 m
Hence, the horizontal distance covered by the shell=1317.4 m
Answer:
72.1 m
Explanation:
Hello!
When the walker walks west, each block he walks will be of 100 m, so the walker walked 4*100m = 400 m west.
Similarly, when the walker walks south, he walks 20 meters per block, therefore, the walker walked 4*20 m = 80 m south.
Since the directions west and south are perpendicular, the distance between the start ad end point is:
d = √(400^2 + 80^2) m = 407.92 m
However the walker traveled 480 m
Therefore, the walker traveled 480 - 407.9 m = 72.1 m farther than the actual distance
Explanation:
Given that,
Mass of the ball, m = 1.2 kg
Initial speed of the ball, u = 10 m/s
Height of the floor from ground, h = 32 m
(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :



v = -25.04 m/s (negative as it rebounds)
The impulse acting on the ball is equal to the change in momentum. It can be calculated as :


J = -42.048 kg-m/s
(b) Time of contact, t = 0.02 s
Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :



F = 0.8409 N
Hence, this is the required solution.
Answer:
160N/m
Explanation:
According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
k is the spring constant
e is the extension
From the formula k = F/e
Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.
The spring constant k = ma/e where
m is the mass of the block = 0.4kg
a is the acceleration = 8.0m/s²
e is the extension of the spring = 2.0cm = 0.02m
K = 0.4×8/0.02
K = 3.2/0.02
K = 160N/m
The spring constant of the spring is therefore 160N/m