Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
A. Average speed is weighted mean (1 × 2 + 2 × 3 + 3 × 5 + 4 × 7 + 3 × 9 + 2 × 12.5)/15 = (2 + 6 + 15 + 28 + 27 + 25)/15 = 103/15 = 6.867 b. RMS is square root of 1/15 times sum of squares of speeds Sum of squares is 4 + 9 + 9 + 25 + 25 + 25 + 49 + 49 + 49 + 49 + 81 + 81 + 81 +156.25 + 156.25 = 848.5
c. RMS speed = √(848.5/15) = 7.521
Most likely the speed is the peak in the speed distribution, which is 7.
Answer:
Weight
Explanation:
The spring balance is used to measure weight of an object.
Answer:
thick wire and cold temperatures
Velocity is the speed and direction combined.
So, the two jets both are going at the same speed.
But they are going in different directions, so their velocities are different.
