As per Newton's II law we know that
![F_{net} = ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20ma)
here we know that
![F_{net} = F_{ap} - F_f](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F_%7Bap%7D%20-%20F_f)
so here we will have
![a = \frac{F_{ap} - F_f}{m}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF_%7Bap%7D%20-%20F_f%7D%7Bm%7D)
so here if we need to increase the acceleration we need to increase the applied force while on increasing the mass or on increasing the friction force the acceleration will decrease.
So here correct answer will be
<em>A) force on the object.</em>
A mass weighing 32 pounds stretches a spring 2 feet.
(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.
(b) How many complete cycles will the mass have completed at the end of 4 seconds?
Answer:
![A = 1.803 ft](https://tex.z-dn.net/?f=A%20%3D%201.803%20ft)
Period =
seconds
8 cycles
Explanation:
A mass weighing 32 pounds stretches a spring 2 feet;
it implies that the mass (m) = ![\frac{w}{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bw%7D%7Bg%7D)
m= ![\frac{32}{32}](https://tex.z-dn.net/?f=%5Cfrac%7B32%7D%7B32%7D)
= 1 slug
Also from Hooke's Law
2 k = 32
k = ![\frac{32}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B32%7D%7B2%7D)
k = 16 lb/ft
Using the function:
![\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2x%7D%7Bdt%7D%20%3D%20-%2016x%5C%5C%5Cfrac%7Bd%5E2x%7D%7Bdt%7D%20%2B%2016x%20%3D0)
(because of the initial position being above the equilibrium position)
( as a result of upward velocity)
NOW, we have:
![x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)](https://tex.z-dn.net/?f=x%28t%29%3Dc_1cos4t%2Bc_2sin4t%5C%5Cx%5E%7B%27%7D%28t%29%20%3D%204%28-c_1sin4t%2Bc_2cos4t%29)
However;
means
![-1 =c_1\\c_1 = -1](https://tex.z-dn.net/?f=-1%20%3Dc_1%5C%5Cc_1%20%3D%20-1)
also implies that:
![-6 =4(c_2)\\c_2 = - \frac{6}{4}](https://tex.z-dn.net/?f=-6%20%3D4%28c_2%29%5C%5Cc_2%20%3D%20-%20%5Cfrac%7B6%7D%7B4%7D)
![c_2 = -\frac{3}{2}](https://tex.z-dn.net/?f=c_2%20%3D%20-%5Cfrac%7B3%7D%7B2%7D)
Hence, ![x(t) =-cos4t-\frac{3}{2} sin 4t](https://tex.z-dn.net/?f=x%28t%29%20%3D-cos4t-%5Cfrac%7B3%7D%7B2%7D%20sin%204t)
![A = \sqrt{C_1^2+C_2^2}](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%7BC_1%5E2%2BC_2%5E2%7D)
![A = \sqrt{(-1)^2+(\frac{3}{2})^2 }](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%7B%28-1%29%5E2%2B%28%5Cfrac%7B3%7D%7B2%7D%29%5E2%20%7D)
![A=\sqrt{\frac{13}{4} }](https://tex.z-dn.net/?f=A%3D%5Csqrt%7B%5Cfrac%7B13%7D%7B4%7D%20%7D)
![A= \frac{1}{2}\sqrt{13}](https://tex.z-dn.net/?f=A%3D%20%5Cfrac%7B1%7D%7B2%7D%5Csqrt%7B13%7D)
![A = 1.803 ft](https://tex.z-dn.net/?f=A%20%3D%201.803%20ft)
Period can be calculated as follows:
= ![\frac{2 \pi}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B2%20%5Cpi%7D%7B4%7D)
=
seconds
How many complete cycles will the mass have completed at the end of 4 seconds?
At the end of 4 seconds, we have:
![x* \frac{\pi}{2} = 4 \pi](https://tex.z-dn.net/?f=x%2A%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%3D%204%20%5Cpi)
![x \pi = 8 \pi](https://tex.z-dn.net/?f=x%20%5Cpi%20%3D%208%20%5Cpi)
cycles
Answer: Transverse
Explanation: Transverse waves possess a vertical wave motion and a horizontal particle motion.
To solve this there is this website that I found that helps
I am in middle school so I have no idea how to solve this
but
this website may help considering u are in high school and u
(hopefully mind u)
know how to solve this
so to get there u google
"whats impact speed"
and click on the first thing there the website is ehow
Answer:
The magnification is -6.05.
Explanation:
Given that,
Focal length = 34 cm
Distance of the image =2.4 m = 240 cm
We need to calculate the distance of the object
![\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bu%7D%2B%5Cdfrac%7B1%7D%7Bv%7D%3D%5Cdfrac%7B1%7D%7Bf%7D)
Where, u = distance of the object
v = distance of the image
f = focal length
Put the value into the formula
![\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bu%7D%3D%5Cdfrac%7B1%7D%7B34%7D-%5Cdfrac%7B1%7D%7B240%7D)
![\dfrac{1}{u}=\dfrac{103}{4080}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bu%7D%3D%5Cdfrac%7B103%7D%7B4080%7D)
![u =\dfrac{4080}{103}](https://tex.z-dn.net/?f=u%20%3D%5Cdfrac%7B4080%7D%7B103%7D)
The magnification is
![m = \dfrac{-v}{u}](https://tex.z-dn.net/?f=m%20%3D%20%5Cdfrac%7B-v%7D%7Bu%7D)
![m=\dfrac{-240\times103}{4080}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B-240%5Ctimes103%7D%7B4080%7D)
![m = -6.05](https://tex.z-dn.net/?f=m%20%3D%20-6.05)
Hence, The magnification is -6.05.