Answer:
Explanation:
a= 7.8i + 6.6j - 7.1k
b= -2.9 i+ 7.4 j+ 3.9k , and
c = 7.6i + 8.8j + 2.2k
r = a - b +c
=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)
= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+ 7.6i + 8.8j + 2.2k
= 18.3 i +18.3 j - k
the angle between r and the positive z axis.
cosθ = 18.3 / √18.3² +18.3² +1
the angle between r and the positive z axis.
= 18.3 / 25.75
cos θ= .71
45 degree
Answer:
Components: 0.0057, -0.0068. Magnitude: 0.0089 m/s
Explanation:
The displacement in the x-direction is:
While the displacement in the y-direction is:
The time taken is t = 304 s.
So the components of the average velocity are:
And the magnitude of the average velocity is
Cody ...
Everything on this page is solved with the SAME formula !
Distance = (speed) x (time) .
Before I get into how to solve each problem, we need to notice that
this whole sheet deals with speed, NOT velocity.
'Velocity' is speed AND THE DIRECTION OF THE MOTION.
Nothing on this page ever mentions direction, so there's no velocity
anywhere on the page.
Your teacher may not be happy if you talk about this on your homework,
but that's too bad. Just don't say "velocity" in any of your answers.
Say "speed", and if the teacher complains about that, then it's time to
let the teacher have it with both barrels.
1). Speed = (distance covered) / (time to cover the distance)
2). Speed = (distance covered) / (time to cover the distance)
3). Distance = (average speed of travel) x (time traveling at that speed)
4). Time to cover the distance = (distance) / (speed)
5). Car's speed = (distance the car covered) / (time the car took)
Sprinter speed = (distance the sprinter covered) / (time the sprinter took)
Calculate the car's speed.
Calculate the sprinter's speed.
... Look at the two speeds.
Decide which one is faster.
... Subtract the slower one from the faster one.
The difference is the answer to "by how much?" .
6). Distance = (speed) x (time spent moving at that speed)
7). Average speed = (TOTAL distance covered)
divided by
(time to cover the TOTAL distance).
