The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Explanation:
Part a)
Angular speed is given as



Part b)
Since turn table is accelerating uniformly
so we will have




Part c)
angular acceleration is given as



Part d)
When its angular speed changes to 120 rpm
then we will have


number of turns revolved is 15 times
so we have



Part e)
now for uniform acceleration we have



<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
Answer:
The electric field strength is 
Solution:
As per the question:
Area of the electrode, 
Charge, q = 50 nC = ![50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m](https://tex.z-dn.net/?f=50%5Ctimes%2010%5E%7B-%209%7D%20C%5B%2Fetx%5D%3C%2Fp%3E%3Cp%3EDistance%2C%20x%20%3D%202%20mm%20%3D%20%5Btex%5D2%5Ctimes%2010%5E%7B-%203%7D%20m)
Now,
To calculate the electric field strength, we first calculate the surface charge density which is given by:

Now, the electric field strength of the electrode is:

where



Answer:
Magnification, m = 3
Explanation:
It is given that,
Focal length of the lens, f = 15 cm
Object distance, u = -10 cm
Lens formula :

v is image distance

Magnification,

So, the magnification of the lens is 3.