Seven
The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer
Eight
They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.
The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read. the difference is 360.
I really do object to the wording, but what can I do?
Nine
Nine is the same thing as 8.
Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408
The actual distance is 400 + 80 = 480
The difference is the answer = 480 - 408 = 72 <<<< Answer
Ten
This is just the displacement magnitude.
dis = sqrt(30^2 + 80^2)
dis = sqrt(900 + 6400)
dis = sqrt(7300)
dis = 85.44 <<<< Answer D
Twelve
Vi = 2.15*Sin(30) = 1.075 m/s
vf = 0
a = - 9.81
t = ?
<u>Formula</u>
a = (vf - vi)/t
<u>Solve</u>
-9.81 = (0 - 1.075)/t
- 9.81 * t = -1.075
t = 0.11 seconds
Thirteen
I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.
Edit
That is a surprise! Really quickly
d = 3.2 m
a = - 9.82
vf = 0
vi = ?
vf^2 = vi^2 - 2*a*d
0 = vi^2 - 2*9.81*3.2
vi = sqrt(19.62*3.2)
vi = 8.0 m/s But that is the vertical component of the speed
v = vi/sin(25)
v = 8.0/sin(25) = 11
To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.
PART A) We will begin by finding the two net distances.
And the distance 'd' is
Through the free-body diagram the tension components are given by
Here we can watch that,
Dividing both expression we have that,
Replacing the values,
PART B) Using the vertical component we can find the tension,
<span>0.0001 km / year or 10^-5 km/year just take 50 km and divide it by 5 million</span>
Answer:
Q=+100kj,w=-15kj,Q=100kj,w=-62kj
Explanation:
when energy is exerted into a system work done is equal to zero .hence the system does work to the surrounding.
