Answer:
C. amount of charge on the source charge.
Explanation:
Electric field lines can be defined as a graphical representation of the vector field or electric field.
Basically, it was first introduced by Michael Faraday and it is typically a curve drawn to the tangent of a point is in the direction of the net field acting on each point.
The number, or density, of field lines on a source charge indicate the amount of charge on the source charge. Therefore, the density of field lines on a source charge is directly proportional to quantity of charge on the source.
The answer is A.
Sy = 1650 x sin30.5 = 837.4 m toward south
Sx = 1650 x cos30.5 = 1421.7 m toward east
The 'net' force acting on the box is (9 - 3) = 6 newtons
in the direction that Carlos is pushing.
Force = (mass) x (acceleration)
6 = (3) x (acceleration)
Divide each side by 3 :
<em>2 m/s² = acceleration</em>
Answer:
C. 110 m/s2
Explanation:
Force = Mass x Acceleration
Since we have the force and the mass, we can rearrange this equation to solve for acceleration by dividing both sides by mass:
Force/Mass = (Mass x Acceleration)/Mass
Acceleration = Force/Mass
Now we just have to plug in our values and calculate!
Acceleration = 48.4/0.44
Acceleration = 110m/s/s
It is option C. 110 m/s2
Hope this helped!
Explanation:
Given that,
Mass of Nichrome, m = 0.5 g
The resistance of the wire, R = 0.673 ohms
Resistivity of the nichrome wire, 
Density, 
(A) The length of the wire is given by using the definition of resistance as :
Volume,

Area,
....(1)

(b) Equation (1) becomes :

Hence, this is the required solution.