Your average speed was
(100 m) / (13.8 s) = 7.25 m/s .
If you finished 0.001s ahead of him, then at your average speed, that corresponds to
(7.25 m/s) x (0.001 s) = 0.00725 m
That's 7.25 millimeters ... about 0.28 of an inch !
NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.
a. reflect (I found this using prior knowledge and process of illumination it can't be absorb cuz you wouldn't see anything it can't be refract because it doesn't reverse the image and it isn't transmit )
Answer:
F = -4567.40 N
Explanation:
Given that,
The power developed by the engine, P = 196 hp
1 hp = 746 W
196 hp = 146157 W
Speed of the car, v = 32 m/s
Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :
F = -4567.40 N
So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.
Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:
(weight) x (distance from the pivot) <u>on one side</u>
is equal to
(weight) x (distance from the pivot) <u>on the other side</u>.
That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.
(Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).
So now we come to the strange beings on the alien planet.
There are three choices right away that both work:
<u>#1).</u>
(400 N) in the middle-seat, facing (200 N) in the end-seat.
(400) x (1) = (200) x (2)
<u>#2).</u>
(200 N) in the middle-seat, facing (100 N) in the end-seat.
(200) x (1) = (100) x (2)
<u>#3).</u>
On one side: (300 N) in the end-seat (300) x (2) = <u>600</u>
On the other side:
(400 N) in the middle-seat (400) x (1) = 400
and (100 N) in the end-seat (100) x (2) = 200
Total . . . . . . . . . . . . <u>600</u>
These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.
