The answer is,
<u>Farthest from the axis of rotation</u>
Please rate <u>Brainliest</u> (:
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)
Work = (force) x (distance)
The work he did: Work = (700 N) x (4m) = 2,800 joules
The rate at which
he did it (power): Work/time = 2,800 joules/2 sec
= 1,400 joules/sec
= 1,400 watts
= 1.877... horsepower (rounded)
Answer
t = 367.77 s = 6.13 min
Explanation:
According to the law of conservation of energy:
![Heat\ Supplied\ By \ Heater = Heat\ Absorbed\ by\ Glass + Heat\ Absorbed\ by\ Water\\Pt = m_gC_g\Delta T_g + m_wC_w\Delta T_w\\](https://tex.z-dn.net/?f=Heat%5C%20Supplied%5C%20By%20%5C%20Heater%20%3D%20Heat%5C%20Absorbed%5C%20by%5C%20Glass%20%2B%20Heat%5C%20Absorbed%5C%20by%5C%20Water%5C%5CPt%20%3D%20m_gC_g%5CDelta%20T_g%20%2B%20m_wC_w%5CDelta%20T_w%5C%5C)
where,
P = Electric Power of Heater = 300 W
t = time required = ?
m_g = mass of glass = 300 g = 0.3 kg
m_w = mass of water = 250 g = 0.25 kg
C_g = speicific heat of glass = 840 J/kg.°C
C_w = specific heatof water = 4184 J/kg.°C
ΔT_g = ΔT_w = Change in Temperature of Glass and water = 100°C - 15°C
ΔT_g = ΔT_w = 85°C
Therefore,
![(300\ W)(t) = (0.3\ kg)(840\ J/kg.^oC)(85^oC)+(0.25\ kg)(4184\ J/kg.^oC)(85^oC)\\](https://tex.z-dn.net/?f=%28300%5C%20W%29%28t%29%20%3D%20%280.3%5C%20kg%29%28840%5C%20J%2Fkg.%5EoC%29%2885%5EoC%29%2B%280.25%5C%20kg%29%284184%5C%20J%2Fkg.%5EoC%29%2885%5EoC%29%5C%5C)
<u>t = 367.77 s = 6.13 min</u>