Answer:
utilization / effects
Explanation:
Utilization equipment are those equipment that makes use of electric energy for the purpose of chemical, electronic, lighting, heating, electro-mechanical or other alike purposes. Hence utilization best suits the first question mark in the question. Secondly, there are associated effects when current flows through a conductor, not responses.
I believe it she should use the first aid kit next
The angle of reflection is "60°".
Here we apply the law of the concept of reflection then we get the final answer easily.
The angle of incident = angle of reflection
Then, the Angle of the incident =60°
What is reflection?
- Reflection is the phenomenon of light rays returning to the source after striking an obstruction.
- It resembles the way a ball bounces when we toss it on a hard surface.
- Some of the light rays that strike an item are reflected, some of them travel through it, and the remainder are absorbed by the object.
- The given values are:Light from a monochromatic source,= 560 nm
- The angle of incidence,= 60°
- The surface of fused quartz (n),= 1.56
- When a light ray does exist on a flat surface, the law or idea of reflection should apply since it includes both the reflected and "normal" light rays at the mirror surface.
- According to the above law,Angle of incident = angle of reflection
- Then, Angle of incident =60°.
To learn more about reflection visit: brainly.com/question/15487308
#SPJ4
Answer:
a)30.14 rad/s2
b)43.5 rad/s
c)60633 J
d)42 kW
e)84 kW
Explanation:
If we treat the propeller is a slender rod, then its moments of inertia is
![I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2](https://tex.z-dn.net/?f=%20I%20%3D%5Cfrac%7BmL%5E2%7D%7B12%7D%20%3D%20%5Cfrac%7B107%2A2.68%5E2%7D%7B12%7D%20%3D%2064.04%20kgm%5E2)
a. The angular acceleration is Torque divided by moments of inertia:
![\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7BT%7D%7BI%7D%20%3D%20%5Cfrac%7B1930%7D%7B64.04%7D%20%3D%2030.14%20rad%2Fs%5E2)
b. 5 revolution would be equals to
rad, or 31.4 rad. Since the engine just got started
![\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5](https://tex.z-dn.net/?f=%5Comega%5E2%20%3D%202%5Calpha%5Ctheta%20%3D%202%2A30.14%2A31.4%20%3D%201893.5)
![\omega = \sqrt{1893.5} = 43.5 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B1893.5%7D%20%3D%2043.5%20rad%2Fs)
c. Work done during the first 5 revolution would be torque times angular displacement:
![W = T*\theta = 1930 * 31.4 = 60633 J](https://tex.z-dn.net/?f=W%20%3D%20T%2A%5Ctheta%20%3D%201930%20%2A%2031.4%20%3D%2060633%20J)
d. The time it takes to spin the first 5 revolutions is
![t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Comega%7D%7B%5Calpha%7D%20%3D%20%5Cfrac%7B43.5%7D%7B30.14%7D%20%3D%201.44%20s)
The average power output is work per unit time
or 42 kW
e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:
or 84 kW
Answer:
im sure your already past this but it's E.
Explanation:
This is because in this case potential energy is linear to height, which means that the higher the more potential energy.