Eccentricity an air compressor of mass 60kg is mounted on an elastic support (i. e., spring) and operates at a speed of 1000 rpm. the compressor is out of balance with an unbalanced mass of 2kg and eccentricity of 0.1m. there is no damping and the spring rate is not known for part a)- write out the equation of motion with all available values b)- if the spring rate of the elastic support is 50 n/mm, what is the magnitude of the displacement? c)-find a new spring rate for the elastic support that results in a transmitted force equal to 25% of the unbalanced load (meω^2).
11.025
Explanation:
Let h = the distance from the edge of the wall to the water surface (m).
Use g = 9.8 m/s² and neglect air resistance.
The initial vertical velocity of the pebble is zero.
Because the pebble hits the surface of the water after 1.5 s, therefore
h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m
S = 11.025 m
Explanation:
Given,
The time taken by the pebble to hit the water surface is, t = 1.5 s
Acceleration due to gravity, g = 9.8 m/s²
Using the II equations of motion
S = ut + 1/2 gt²
Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity
u = 0
Therefore, the equation becomes
S = 1/2 gt²
Substituting the given values in the above equation
S = 0.5 x 9.8 x 1.5²
= 11.025 m
Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m
Answer is below. Hope this helps you!
Step-by-step explanation:
(i)v = u + at
(ii) s = ut + 1/2 *at²
(iii)2as = v²-u²
Examples:
A train starting from rest attains a velocity of 72km/h in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
72km/h = 20m/s
(i) use v= u+ at
20 = 0 + a*300
20 = 5a
20 / 300= a
1/15 ms² = a
to find distance
use 2as= v²-u²
2 *1/15 * s = (20) - 0²
2/15s = 400
s = 400 * 15/2
s = 3000m = 3 km
a car accelerates with a uniform acceleration of 1m/s². it starts with a velocity of 18km/h and reaches 36 km/h in 5 seconds find the distance travelled.
18 km /h = 5m/s
36km/h = 10m/s
use - s= ut + 1/2at²
s = 5 * 5 + 1/2 * 1 * (5)²
s = 25 + 25/2
s = 37.5 meters
Let h = the distance from the edge of the wall to the water surface (m).
Use g = 9.8 m/s² and neglect air resistance.
The initial vertical velocity of the pebble is zero.
Because the pebble hits the surface of the water after 1.5 s, therefore
h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m
11.025 m
It is described in terms of displacement, distance, velocity, acceleration, time and speed. Jogging, driving a car, and even simply taking a walk are all everyday examples of motion.
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The three equations are,
v = u + at.
v² = u² + 2as.
s = ut + ½at²