Answer:
1) a block going down a slope
2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK
Explanation:
In this exercise you are asked to give an example of various types of systems
1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.
2)
a) rolling a ball uphill
In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact
W = ΔU + ΔK + ΔE
b) in this system work is transformed into internal energy
W = ΔE
c) There is no friction here, therefore the work is transformed into kinetic energy
W = ΔK
d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy
ΔU = ΔK
Answer:
![a=368.97\ m/s^2](https://tex.z-dn.net/?f=a%3D368.97%5C%20m%2Fs%5E2)
Explanation:
Given that,
Initial angular velocity, ![\omega=0](https://tex.z-dn.net/?f=%5Comega%3D0)
Acceleration of the wheel, ![\alpha =7\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D7%5C%20rad%2Fs%5E2)
Rotation, ![\theta=14\ rotation=14\times 2\pi =87.96\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D14%5C%20rotation%3D14%5Ctimes%202%5Cpi%20%3D87.96%5C%20rad)
Let t is the time. Using second equation of kinematics can be calculated using time.
![\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_it%2B%5Cdfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2%5C%5C%5C%5Ct%3D%5Csqrt%7B%5Cdfrac%7B2%5Ctheta%7D%7B%5Calpha%20%7D%7D%20%5C%5C%5C%5Ct%3D%5Csqrt%7B%5Cdfrac%7B2%5Ctimes%2087.96%7D%7B7%7D%7D%20%5C%5C%5C%5Ct%3D5.01%5C%20s)
Let
is the final angular velocity and a is the radial component of acceleration.
![\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Calpha%20t%5C%5C%5C%5C%5Comega_f%3D0%2B7%5Ctimes%205.01%5C%5C%5C%5C%5Comega_f%3D35.07%5C%20rad%2Fs)
Radial component of acceleration,
![a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Comega_f%5E2r%5C%5C%5C%5Ca%3D%2835.07%29%5E2%5Ctimes%200.3%5C%5C%5C%5Ca%3D368.97%5C%20m%2Fs%5E2)
So, the required acceleration on the edge of the wheel is
.
Answer:
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Answer:
Hyoid
Explanation:
The hyoid is located in the neck area, not the limbs.