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
The formula that can be used to obtain a convergent series is given by;
a/1 – r.
<h3>What is a series?</h3>
In mathematics, we define a series as sum of numbers which could be convergent or divergent. In a convergent series, the summation of the numbers approaches a given value.
The formula that can be used to obtain a convergent series is given by;
a/1 – r.
Learn more about a convergent series:brainly.com/question/15415793?
#SPJ11
Answer:
7. They arethe meter (m), the kilogram (kg), the second (s), the kelvin (K), the ampere (A), the mole (mol), and the candela (cd)
Explanation:
7. They arethe meter (m), the kilogram (kg), the second (s), the kelvin (K), the ampere (A), the mole (mol), and the candela (cd)
Answer:
13.5
Explanation:
Mass: 5kg
Initial Velocity: -15
Final Velocity: 12
Force: 10
We can use the equation: Vf = Vi + at
We need to find acceleration, and we can use the equation, F=ma,
We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.
Now we have all the variables to find time.
Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)
Plugging them in into desmos gives 13.5 for time.