I really don’t know but the question is confusing can you explain more!!!
Answer:
Respect is having a regard for other people and their lives; it is showing those around us compassion and empathy. Children who show respect will find they are successful in all aspects of life. The sports environment is a great place to grow and establish respect.
What type of reaction is being shown in this energy diagram
Answer:
x = 5.29 m
Explanation:
given,
weight of stretch = 37 N
left-hand spring constant (k₁)= 2.7 N/cm
right hand spring constant(k₂)= 4.3 N/ cm
spring are connected in parallel
F = F₁ + F₂
F = k₁x + k₂x
F = (k₁+ k₂)x
37= (4.3+ 2.7)x
7 x = 37
x = 5.29 m
Answer:
(a) ![\alpha=-111.26rad/s](https://tex.z-dn.net/?f=%5Calpha%3D-111.26rad%2Fs)
(b) ![s=4450.6in](https://tex.z-dn.net/?f=s%3D4450.6in)
(c) ![8.66in](https://tex.z-dn.net/?f=8.66in)
Explanation:
First change the units of the velocity, using these equivalents
and ![1 min =60s](https://tex.z-dn.net/?f=1%20min%20%3D60s)
![4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s](https://tex.z-dn.net/?f=4250rpm%28%5Cfrac%7B2%5Cpi%20rad%7D%7B1rev%7D%29%28%5Cfrac%7B1%20min%7D%7B60%20s%7D%20%29%3D445.06rad%2Fs)
The angular acceleration
the time rate of change of the angular speed
according to:
![\alpha=\frac{\Delta \omega}{\Delta t}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7B%5CDelta%20%5Comega%7D%7B%5CDelta%20t%7D)
![\Delta \omega=\omega_i-\omega_f](https://tex.z-dn.net/?f=%5CDelta%20%20%5Comega%3D%5Comega_i-%5Comega_f)
Where
is the original velocity, in the case the velocity before starting the deceleration, and
is the final velocity, equal to zero because it has stopped.
![\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7B%5CDelta%20%5Comega%7D%7B%5CDelta%20t%7D%20%3D%5Cfrac%7B%5Comega_i-%5Comega_f%7D%7B4%7D%5Cfrac%7B0-445.06%7D%7B4%7D%20%3D%5Cfrac%7B-445.06%7D%7B4%7D%20%3D-111.26rad%2Fs)
b) To find the distance traveled in radians use the formula:
![\theta = \omega_i t + \frac{1}{2} \alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega_i%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5Calpha%20t%5E2)
![\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20445.06%20%284%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28-111.26%29%20%284%29%5E2%3D1780.24-890.12%3D890.12rad)
To change this result to inches, solve the angular displacement
for the distance traveled
(
is the radius).
![\theta=\frac{s}{r} \\s=\theta r](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7Bs%7D%7Br%7D%20%5C%5Cs%3D%5Ctheta%20r)
![s=890.12(5)=4450.6in](https://tex.z-dn.net/?f=s%3D890.12%285%29%3D4450.6in)
c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:
![\frac{890.12}{2\pi}=141.6667](https://tex.z-dn.net/?f=%5Cfrac%7B890.12%7D%7B2%5Cpi%7D%3D141.6667)
The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle
is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):
![c^2=a^2+b^2-2abcos(\gamma)](https://tex.z-dn.net/?f=c%5E2%3Da%5E2%2Bb%5E2-2abcos%28%5Cgamma%29)
![c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in](https://tex.z-dn.net/?f=c%5E2%3D5%5E2%2B5%5E2-2%285%29%285%29cos%28%5Cfrac%7B2%5Cpi%7D%7B3%7D%20%29%5C%5Cc%5E2%3D25%2B25%2B25%5C%5Cc%5E2%3D75%5C%5Cc%3D5%5Csqrt%7B3%7D%3D8.66in)