<em>F</em> = 153 N
Explanation:
Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components 
and 
:
= 350 N - 200 N = 150 N
= 180 N - 150 N = 30 N
The magnitude of the resultant force <em>F</em> is given by
To find the direction 
, we use
or
about 5 watts (5W) of power
The total capacitance is <em>C</em> such that
1/<em>C</em> = 1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)
Solve for <em>C</em> :
<em>C</em> = 1 / (1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)) ≈ 3.1 µF
Answer:
150m
Explanation:
The relation of speed/time and distance/time is a derivative/integral one, as in speed is the derivative of distance (the faster you go, the faster the distance changes, duh!).
So we need to compute the integral of speed over time from 0.0s to 5.0s.
The easiest way here is to compute the area under the line (it's going to be faster than computing the acceleration and using a formula of distance based on acceleration).
The area under the line is a trapezoid with "height" 5s, and the bases 10m/s and 50m/s. Using the trapezoid area formula of h*(a + b)/2
distance = 5s * (10m/s + 50m/s) / 2 = 5s * 60m/s / 2 = 5s * 30m/s = 150m
Alternatively, we can use the acceleration formula:
a = (50m/s - 10m/s)/5s = 40m/s / 5s = 8m/s^2
distance = v0 * t + a * t^2 / 2 = 10m/s * 5s + 8m/s^2 * (5s)^2 / 2 = 50m + 8m * 25 / 2 = 50m + 100m = 150m.
