Answer:
(35 N - 10 N)/8kg = 3.125 m/s^2
Explanation:
The formula for Force is:
Force = Mass*Acceleration
(Force is equal to Mass times Acceleration)
Since we're told to find the acceleration of the box. We make acceleration the subject of the equation:
Acceleration = Force/Mass
(Acceleration equal to Force divided by Mass)
We know that the force are 35 N forward and 10 N backward, and the weight of the box is 8kg.
= (35 N - 10 N)/8kg
The reason that 35 N minus 10 N is because the 10 N is pushing the box backward.
= 25 N/8kg
= 3.125 m/s^2
Hope it helps :DD
In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>
Answer:
(a) 1.58 V
(b) 0.0126 Wb
(c) 0.0493 V
Solution:
As per the question:
No. of turns in the coil, N = 400 turns
Self Inductance of the coil, L = 7.50 mH = 
Current in the coil, i = ![1680cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=1680cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
A
where
Now,
(a) To calculate the maximum emf:
We know that maximum emf induced in the coil is given by:
![e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20L%5Cfrac%7Bd%7D%7Bdt%7D%281680%29cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
![e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20-%207.50%5Ctimes%2010%5E%7B-%203%7D%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B0.0250%7D%5Ctimes%20%5Cfrac%7Bd%7D%7Bdt%7D%281680%29sin%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
For maximum emf, 
should be maximum, i.e., 1
Now, the magnitude of the maximum emf is given by:
(b) To calculate the maximum average flux,we know that:
(c) To calculate the magnitude of the induced emf at t = 0.0180 s:
