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

4 0

3 years ago

Is (2, 2) a solution of y < 4x-6<br> Help

vesna_86 [32]

Answer:

B) No.

Explanation:

Okay,so,

this is equation is y=mx +b

mx represents the slope

and b represents the y-intercept

in order to figure this out you need to plot the y-intercept first

that makes its (0,-6) because the 6 is negative in the equation

4x is also equal to 4/1 since we dont know what x is

we have to do rise over run for this

you go up 4 spots on the y intercept from -6 because 4 is positive

then you go to the right 1 time because 1 is positive.

this leaves you at (1,-2)

so, (2,2) is NOT a solution

7 0

3 years ago

What is measurement​

What is measurement