Answer:
Explanation:
In a L C R circuit, the average power is given by
![P_{av}=V_{rms}I_{rms}Cos\phi](https://tex.z-dn.net/?f=P_%7Bav%7D%3DV_%7Brms%7DI_%7Brms%7DCos%5Cphi)
As given in the question
CosФ = R / Z
And we know that
![V_{rms}=I_{rms}\times Z](https://tex.z-dn.net/?f=V_%7Brms%7D%3DI_%7Brms%7D%5Ctimes%20Z)
So
![P_{av}=I_{rms}\times Z\times I_{rms}\times \frac{R}{Z}](https://tex.z-dn.net/?f=P_%7Bav%7D%3DI_%7Brms%7D%5Ctimes%20Z%5Ctimes%20I_%7Brms%7D%5Ctimes%20%5Cfrac%7BR%7D%7BZ%7D)
![P_{av}=I_{rms}^{2}\times R](https://tex.z-dn.net/?f=P_%7Bav%7D%3DI_%7Brms%7D%5E%7B2%7D%5Ctimes%20R)
In physical chemistry, the terms body-centered cubic (BCC) and face-centered cubic (FCC) refer to the cubic crystal system of a solid. Each solid is made up simple building blocks called lattice units. There are different layouts of a lattice unit.
It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.
Answer:
0.42 m/s²
Explanation:
r = radius of the flywheel = 0.300 m
w₀ = initial angular speed = 0 rad/s
w = final angular speed = ?
θ = angular displacement = 60 deg = 1.05 rad
α = angular acceleration = 0.6 rad/s²
Using the equation
w² = w₀² + 2 α θ
w² = 0² + 2 (0.6) (1.05)
w = 1.12 rad/s
Tangential acceleration is given as
= r α = (0.300) (0.6) = 0.18 m/s²
Radial acceleration is given as
= r w² = (0.300) (1.12)² = 0.38 m/s²
Magnitude of resultant acceleration is given as
![a = \sqrt{a_{t}^{2} + a_{r}^{2}}](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7Ba_%7Bt%7D%5E%7B2%7D%20%2B%20a_%7Br%7D%5E%7B2%7D%7D)
![a = \sqrt{0.18^{2} + 0.38^{2}}](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7B0.18%5E%7B2%7D%20%2B%200.38%5E%7B2%7D%7D)
= 0.42 m/s²
To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.
Centrifugal acceleration is determined by
![a_c = \frac{v^2}{r}](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D)
Where,
v = Velocity
r = Radius
From the given data of the moon we know that gravity there is equivalent to
![a = 1.62m/s](https://tex.z-dn.net/?f=a%20%3D%201.62m%2Fs)
While the radius of the moon is given by
![r = 1.74*10^6m](https://tex.z-dn.net/?f=r%20%3D%201.74%2A10%5E6m)
If we rearrange the function to find the speed we will have to
![v = \sqrt{ar}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bar%7D)
![v = \sqrt{1.6(1.74*10^6)}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B1.6%281.74%2A10%5E6%29%7D)
![v = 1.7km/s](https://tex.z-dn.net/?f=v%20%3D%201.7km%2Fs)
The speed for this to happen is 1.7km/s