Answer:
D
Step-by-step explanation:
Look at the file i attached
You didn't include the formula.
Given that there is no data about the mass, I will suppose that the formula is that of the simple pendulum (which is only valid if the mass is negligible).
Any way my idea is to teach you how to use the formula and you can apply the procedure to the real formula that the problem incorporates.
Simple pendulum formula:
Period = 2π √(L/g)
Square both sides
Period^2 = (2π)^2 L/g
L = [Period / 2π)^2 * g
Period = 3.1 s
2π ≈ 6.28
g ≈ 10 m/s^2
L = [3.1s/6.28]^2 * 10m/s^2 =2.43 m
I hope this helps you.
Answer:
Step-by-step explanation:
Sum of two sides of triangle always greater than third side, so the possible values of h is:
7x-3x < h < 7x+3x
4x < h < 10x
The answer to this question is:
A circle is growing so that the radius is increasing at the rate of 2cm/min. How fast is the area of the circle changing at the instant the radius is 10cm? Include units in your answer.?
✔️I assume here the linear scale is changing at the rato of 5cm/min
✔️dR/dt=5(cm/min) (R - is the radius.... yrs, of the circle (not the side)
✔️The rate of area change would be d(pi*R^2)/dt=2pi*R*dR/dt.
✔️At the instant when R=20cm,this rate would be,
✔️2pi*20*5(cm^2/min)=200pi (cm^2/min) or, almost, 628 (cm^2/min)
Hoped This Helped, <span>Cello10
Your Welcome :) </span>
