Answer:
The work done is -209.42 J.
Explanation:
F(x) = (- 20 - 3 x ) N
x = 0 to x = 6.9 m
Here, the force is variable in nature, so the work done by the variable force is given by
![W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J](https://tex.z-dn.net/?f=W%20%3D%5Cint%20F%20dx%5C%5C%5C%5CW%20%3D%5Cint_%7B0%7D%5E%7B6.9%7D%28-20-%203x%20dx%20%29%5C%5C%5C%5CW%3D%20%5Cleft%20%5B%20-%2020%20x%20-%201.5%20x%5E2%20%5Cright%20%5D_%7B0%7D%5E%7B6.9%7D%5C%5C%5C%5CW%20%3D%20-%2020%20%286.9%20-%200%29%20-%201.5%286.9%5Ctimes%206.9%20-%200%29%5C%5C%5C%5CW%20%3D-%20138%20-%2071.42%5C%5C%5C%5CW%20%3D%20-%20209.42%20%20J)
Say the initial point is (0,0)
The final point is
x = 200 + 135*cos(30) = 200 + 135*sqrt(3)/2 = 316.91 ft
y = 135*sin(30) = 135/2 = 67.5 ft
Resultant vector = (316.91, 67.5) - (0,0) = 316.91, 67.5) ft
<span>5.98 x 10^-2 ohms.
Resistance is defined as:
R = rl/A
where
R = resistance in ohms
r = resistivity (given as 1.59x10^-8)
l = length of wire.
A = Cross sectional area of wire.
So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives:
R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2)
R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7)
R = (4.77 x 10^-8) / (7.98015 x 10^-7)
R = 5.98 x 10^-2 ohms
So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in the direction that the charge is moving? The answer is <span>thumb.
</span>One way to remember this is that there is one velocity, represented accordingly by the thumb. There are many field lines, represented accordingly by the fingers. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge. Because the force is always perpendicular to the velocity vector, a pure magnetic field will not accelerate a charged particle in a single direction, however will produce circular or helical motion (a concept explored in more detail in future sections). It is important to note that magnetic field will not exert a force on a static electric charge. These two observations are in keeping with the rule that <span>magnetic fields do no </span>work<span>.</span>
