Answer:
25 N
Explanation:
Work is a product of force and perpendicular distance moved.
W=Fd where F is force exerted and d is perpendicular distance.
However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as 
Therefore, 
Making F the subject of the formula then
where
is the angle of inclination. Substituting 190 J for W then 18 degrees for
and 8 m for d then
Since each time trial is the same the average will be the direct answer, and the formula for velocity is distance divided by time, therefore it will come out to, 1.92307692. Whatever your teacher what the rounding process to be will vary but the straight up answer is there.
Answer:
Displacement: 6.71 m, Direction: 63.4 degrees north of east
Explanation:
In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).
Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).
And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)
And the displacement will be defined by the folliwing vector operation:

We know that the magnitude of the displacement vector is defined by the phytagoras theorem

And the angle will be defined by:
tan(beta)=3/6
beta = tan^-1(6/3)
beta = 63.43°

P = Power
W = Work
Delta T = Change of Time
So:
P = 35 Joules
Delta T = 70 secs
W = ?

* Joule/Sec = Watt
Answer: The power used to do this task was 0.5 Watts.
Answer:
In the case of a negatively charged rod, the field lines points radially towards the rod.
Explanation:
Field lines pointing towards the negatively charged rod are used to represent the nature of an electric field. Field lines always points in a direction, i.e If the charge is positive, field lines points radially away from the rod and if the charge is negative the field lines points radially towards the rod. So in the case of a negatively charged rod, the field lines points radially towards the rod.