Answer:
FB = 0.187 N
Explanation:
To find the magnetic force FB in the wire you use the following formula:

the angle between B and L is given by:

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:
![|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N](https://tex.z-dn.net/?f=%7C%5Cvec%7BF_B%7D%7C%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7DB%28y%29dy%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7D%280.5y%29dy%5C%5C%5C%5C%7C%5Cvec%7BF_B%7D%7C%3D%282.0%2A10%5E%7B-3%7DA%29%28sin36.86%5C%C2%B0%29%280.5T%29%5B%5Cfrac%7B0.25%5E2%7D%7B2%7Dm%5D%3D0.187%5C%20N)
hence, the magnitude of the magnetic force is 0.187N
Answer:
If resistance increases current decreases.
Explanation:
- Current is <em>inversely proportional</em> to the resistance.
- from the relation given below, we can clearly see the relation between current and resistance;
V=IR
I ∝ 1/R
This relation shows that when resistance increases,current decreases.
Answer:
Part a)

Part b)

Explanation:
As we know that the see saw bar is massless so here torque due to two masses is given as

here we will have

now we will have inertia of two masses given as

now we have

now the angular acceleration is given as

so we have

Part b)
Now if the rod is not massles then we will have total inertia given as

so we will have

now the acceleration is given as


In order to create a charged object you need to transfer electrons either away or to the object by induction, conduction, or friction
Induction is without contact(like bringing a charged object to a electroscope charges the leaves at the bottom)
Conduction is with contact(like the previous answer, wore touching transfers the charge from a source to the object)
Friction(rubbing a balloon on wool)
Answer:
19.01 N
Explanation:
F = Force being applied to the crate = 45 N
= Angle at which the force is being applied = 
Horizontal component of force is given by

The horizontal component of the force acting on the crate is 19.01 N.