Hence the expression of ω in terms of m and k is
![\omega = \sqrt{\frac{k}{m}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D)
Given the expressions;
![T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}](https://tex.z-dn.net/?f=T_s%20%3D%202%20%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7D%20%5C%20and%20%5C%20T_s%20%3D%20%5Cfrac%7B2%20%5Cpi%7D%7B%5Comega%7D)
Equating both expressions we will have;
![2 \pi \sqrt{\frac{m}{k} } = \frac{2 \pi}{\omega}](https://tex.z-dn.net/?f=2%20%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7D%20%20%3D%20%5Cfrac%7B2%20%5Cpi%7D%7B%5Comega%7D)
Divide both equations by 2π
![\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\](https://tex.z-dn.net/?f=%5Cfrac%7B2%20%5Cpi%5Csqrt%7B%5Cfrac%7Bm%7D%7B2%20%5Cpi%7D%20%7D%20%7D%7B2%20%5Cpi%7D%3D%5Cfrac%7B%5Cfrac%7B2%20%5Cpi%7D%7B%5Comega%7D%20%7D%7B2%5Cpi%7D%5C%5C%5Csqrt%7B%5Cfrac%7Bm%7D%7B2%20%5Cpi%7D%20%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Comega%7D%5C%5C)
Square both sides
![(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}](https://tex.z-dn.net/?f=%28%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7D%20%29%5E2%20%3D%20%28%5Cfrac%7B1%7D%7B%5Comega%7D%20%29%5E2%5C%5C%5Cfrac%7Bm%7D%7Bk%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Comega%20%5E2%7D%20%5C%5C%5Comega%20%5E2%20%3D%20%5Cfrac%7Bk%7D%7Bm%7D)
Take the square root of both sides
![\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Comega%20%5E2%7D%20%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%20%5C%5C%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D)
Hence the expression of ω in terms of m and k is
![\omega = \sqrt{\frac{k}{m}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D)
Answer:
stryo: 1
wood: 1
ice: 1
brick: 2
aluminum: 2.7
Explanation:
d= mass/ total volume
(fyi: for aluminum, they did the subtraction wrong to find the total volume. it is actually 5 or 5.00)
We must remember that the total net force equation at
constant velocity is:
<span>F – Ff = 0</span>
of
F - µN = 0
Using Newton's 2nd Law of Motion:<span>
F = m a
<span>Where,
F = net force acting on the body
m = mass of the body
a = acceleration of the body
Since the cart is moving at a constant velocity, then
acceleration is zero, hence the working equation simplifies to
F = net Force = 0
Therefore,
F - µN = 0
where
µ = coefficient of friction = 0.20
N = normal force acting on the cart = 12 N
Therefore,
F - 0.20(12) = 0
<span>
F = 2.4 N </span></span></span>
The amount left of a given substance can be calculated through the equation,
A = (A0) x 0.5^n/h
From the given scenario,
A/A0 = 0.75 = 0.5*(60/h)
The value of h from the equation is 144.565 minutes.