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Answer:
Law of Motion/ Law of Inertia
Explanation:
Objects at rest stay at rest unless you're the one to move it.
Answer:
640 J
Explanation:
At constant pressure, work = pressure * change in volume.
First, let's convert to SI units.
200 kPa = 200,000 Pa
3.4 L - 0.2 L = 3.2 L = 0.0032 m³
Therefore:
W = (200,000 Pa) (0.0032 m³)
W = 640 J
Answer:
The time is
Explanation:
From the question we are told that
The diameter of the circular saw is ![d = 10 \ in = \frac{10}{12} = 0.833 \ feet](https://tex.z-dn.net/?f=d%20%3D%2010%20%5C%20in%20%3D%20%5Cfrac%7B10%7D%7B12%7D%20%20%3D%200.833%20%5C%20feet)
The peripheral speed is ![u = 230 \ ft/s](https://tex.z-dn.net/?f=u%20%3D%20230%20%5C%20ft%2Fs)
The time taken for the blade to come to rest is t = 17 s
The total acceleration of the tooth considered is
Generally the radius of the blade is mathematically represented as
![r = \frac{0.833}{2}= 0. 4165 \ feet](https://tex.z-dn.net/?f=r%20%3D%20%20%5Cfrac%7B0.833%7D%7B2%7D%3D%200.%204165%20%20%20%20%5C%20feet)
Generally the tangential acceleration of the blade is mathematically represented as
![a__{t}} = \frac{v - u}{t}](https://tex.z-dn.net/?f=a__%7Bt%7D%7D%20%3D%20%5Cfrac%7Bv%20-%20u%7D%7Bt%7D)
Here v is the final velocity of the tooth of the blade which is zero since the blade came to rest
so
![a__{t}} = \frac{0 - 230}{ 17}](https://tex.z-dn.net/?f=a__%7Bt%7D%7D%20%3D%20%5Cfrac%7B0%20-%20230%7D%7B%2017%7D)
=> ![a__{t}} = - 13.53 \ ft/s^2](https://tex.z-dn.net/?f=a__%7Bt%7D%7D%20%3D%20-%2013.53%20%5C%20ft%2Fs%5E2)
Generally the total acceleration of the tooth of the blade is mathematically represented as
![a = \sqrt{a_t^2 + a_r^2}](https://tex.z-dn.net/?f=a%20%20%3D%20%5Csqrt%7Ba_t%5E2%20%2B%20a_r%5E2%7D)
Here
is the radial acceleration , now making
the subject of the formula we have that
![130= \sqrt{13.56 ^2 + a_r^2}](https://tex.z-dn.net/?f=130%3D%20%5Csqrt%7B13.56%20%5E2%20%2B%20a_r%5E2%7D)
=> ![a_r = \sqrt{130^2 -(- 13.56)^2}](https://tex.z-dn.net/?f=a_r%20%3D%20%20%5Csqrt%7B130%5E2%20-%28-%2013.56%29%5E2%7D)
=> ![a_r = 129.3 \ m/s^2](https://tex.z-dn.net/?f=a_r%20%3D%20%20129.3%20%5C%20m%2Fs%5E2)
Generally radial acceleration is mathematically represented as
Here
is the velocity at which the total acceleration is 130 ft/s2.
=>
=> ![v_r = \sqrt{129.3 * 0.4165 }](https://tex.z-dn.net/?f=v_r%20%3D%20%5Csqrt%7B129.3%20%2A%200.4165%20%7D)
=> ![v_r = 7.34 \ m/s](https://tex.z-dn.net/?f=v_r%20%3D%207.34%20%5C%20m%2Fs)
Generally the time at which the total acceleration is 130 ft/s2. is mathematically represented as
![t_r = \frac{7.34 - 300}{a_t}](https://tex.z-dn.net/?f=t_r%20%3D%20%5Cfrac%7B7.34%20%20-%20300%7D%7Ba_t%7D)
=> ![t_r = \frac{7.34 - 300}{-13.56}](https://tex.z-dn.net/?f=t_r%20%3D%20%5Cfrac%7B7.34%20-%20300%7D%7B-13.56%7D)
=>
Answer:
The magnitude and direction of its angular momentum relative to point is 115.28 kg m²/s.
Explanation:
Given that,
Mass = 2.00 kg
Velocity = 12.0 m/s
Suppose the distance from point O to point p and angle is 8.00 m and 36.9°.
We need to calculate the magnitude and direction of its angular momentum relative to point
Using formula of angular momentum
![L=mvr\sin\theta](https://tex.z-dn.net/?f=L%3Dmvr%5Csin%5Ctheta)
Where, r = distance from point O to point P
v = speed
= direction
![L=mvr\sin(180-\alpha)](https://tex.z-dn.net/?f=L%3Dmvr%5Csin%28180-%5Calpha%29)
Put the value into the formula
![L=2.00\times12.0\times8.00\sin(180-36.9)](https://tex.z-dn.net/?f=L%3D2.00%5Ctimes12.0%5Ctimes8.00%5Csin%28180-36.9%29)
![L=115.28\ kg\ m^2/s](https://tex.z-dn.net/?f=L%3D115.28%5C%20kg%5C%20m%5E2%2Fs)
The direction of angular momentum is pointing into the page.
Hence, The magnitude and direction of its angular momentum relative to point is 115.28 kg m²/s.