Answer:
The invention of the pendulum-driven ___<u>clocks</u>___ in the 1600s paved the way for a new industrial era.
Answer:
80grit
Explanation:
80 grit is coarsest grit that may be used on aluminum
The lowest grit sizes range from 40 to 60. From the given options 80 grit is practically available grit.
What is a sandpaper used for?
They are essentially used for surface preparation. Sandpaper is produced in a range of grit sizes and is used to remove material from surfaces, either to make them smoother (for example, in painting and wood finishing), to remove a layer of material (such as old paint), or sometimes to make the surface rougher (for example, as a preparation for gluing).
Answer:
radius = 0.045 m
Explanation:
Given data:
density of oil = 780 kg/m^3
velocity = 20 m/s
height = 25 m
Total energy is = 57.5 kW
we have now
E = kinetic energy+ potential energy + flow work
![E = \dot m ( \frac{v^2}{2] + zg + p\nu)](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%20%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p%5Cnu%29)
![E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p_%7Batm%7D%20%5Cfrac%7B1%7D%7B%5Crho%7D%29)

solving for flow rate
![\dot m = 99.977we know that [tex]\dot m = \rho AV](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%2099.977%3C%2Fp%3E%3Cp%3Ewe%20know%20that%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cdot%20m%20%20%3D%20%5Crho%20AV)

solving for d

d = 0.090 m
so radius = 0.045 m
Answer:
V = 280.15 V
Explanation:
" The complete question is attached with figure"
Given:
- The capacitance of the capacitor C = 10 nF
- The amount of mass attached to motor m = 4 grams
- The amount of distance it is to be lifted h = 1 cm
- Ignore all other losses in the system
Find:
- The voltage required to lift the mass m through distance h?
Solution:
- The conservation of energy for the entire system is written as:
Work_gravity = U_c
Where,
Work_gravity: Work done by gravity on mass m
U_c: The amount of energy stored in a capacitor
m*g*h = 0.5*C*V^2
V^2 = 2*m*g*h / C
V = sqrt ( 2*m*g*h / C )
Plug in the values:
V = sqrt ( 2*0.004*9.81*0.01 / 10*10^-9 )
V = sqrt ( 78,480)
V = 28.15 V