Answer:
Incomplete question, check attachment for completed question
Explanation:
The force of attraction between two forces are given as
F=kQq/r²
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
If he feels like, is interested in it, and is able to grasp it, then why not ? Why not indeed ?
Number of miles that marker shows when passes through town= 160 miles.
Number of miles that marker shows currently to John = 115 miles.
We need to find the distance between town and John's current location.
For the problem, we can clearly see that Town is at 160 miles away but when John passes the marker shows 115 miles.
So, it's just the difference between 160 miles and 115 miles.
In order to find that difference, we need to subtract those two numbers.
160miles - 115miles = 45 miles.
So, we could say the distance between town and John's current location is 45 miles.
