Answer:
The magnitude of angular acceleration is 
.
Explanation:
Given that,
Initial angular velocity, 
When it switched off, it comes o rest, 
Number of revolution, 
We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :
So, the magnitude of angular acceleration is . Hence, this is the required solution.
Answer:........u are so sweet
Answer:
a) 360 kQ
b) 4.32 MJ
c) 1200 W-h
Explanation:
a) The definition of the current is.
having a count that 10 hours = 36.000 s
![Q = 36000*10 = 360.000 [Q]](https://tex.z-dn.net/?f=%20Q%20%3D%2036000%2A10%20%3D%20360.000%20%5BQ%5D%20)
b) The definition of the Energy in power terms is.
and the definition of power is:
![P = V*I = 12 * 10 = 120 [W]](https://tex.z-dn.net/?f=%20P%20%3D%20V%2AI%20%3D%2012%20%2A%2010%20%3D%20120%20%5BW%5D)
replacing in the energy formula.
solving the integral, have into account that t is in seconds.
![E=P*t=120*36000=4.320.000=4.32 [MJ]](https://tex.z-dn.net/?f=%20E%3DP%2At%3D120%2A36000%3D4.320.000%3D4.32%20%5BMJ%5D)
c) The energy in W-h, we can find it multiplying power by hours
.
<span>a. The magnitude of the vector is doubled as well.
Let's say we have a 2-dimensional vector with components x and y.
It's magnitude lâ‚ is given by:
lâ‚ = âš(x² + y²)
If we double the components x and y, the new magnitude lâ‚‚ is:
lâ‚‚ = âš((2x)² + (2y²))
With a bit of algebra...
lâ‚‚ = âš(4x² + 4y²)
lâ‚‚ = âš4(x² + y²)
lâ‚‚ = 2âš(x² + y²)
We can write the new magnitude lâ‚‚ in terms of the old magnitude lâ‚.
lâ‚‚ = 2lâ‚
Therefore, the new magnitude is double the old one.
It should be clear that this relationship applies to 3D (and 1D) vectors as well.
b. The direction angle is unchanged.
The direction angle θ₠for a 2-dimensional vector is given by:
θ₠= arctan(y / x)
If we double both components, we get:
θ₂ = arctan(2y / 2x)
θ₂ = arctan(y / x)
θ₂ = θâ‚
The new direction angle is the same as the old one.</span>
