Answer:
a. by moving the book without acceleration and keeping the height of the book constant
Explanation:
FOR CONSTANT KINETIC ENERGY:
The kinetic energy of a body depends upon its speed according to its formula:
ΔK.E = (1/2)mΔv²
So, for Δv = 0 m/s
ΔK.E = 0 J
So, for keeping kinetic energy constant, the books must be moved at constant speed without acceleration.
FOR CONSTANT POTENTIAL ENERGY:
The potential energy of a body depends upon its height according to its formula:
ΔP.E = mgΔh
So, for Δh = 0 m/s
ΔP.E = 0 J
So, for keeping potential energy constant, the books must be moved at constant height.
So, the correct option is:
<u>a. by moving the book without acceleration and keeping the height of the book constant</u>
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
Ok well I know measure of long leg is 30 degrees and short leg is 60 degrees
Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?
Answer:
601000 N
Explanation:
Given that :
Acceleration due to gravity at lunar outpost = 1.6m/s²
Supported Weight of supplies = 1 * 10^5 N
Acceleration due to gravity on the earth surface = 9.8m/s²
Maximum weight of supplies as measured on EARTH :
Ratio of earth gravity to lunar post gravity:
(Earth gravity / Lunar post gravity) ;
(9.8 / 1.63) = 6.01
Hence, maximum weight of supplies as measured on EARTH should be :
6.01 * (1 × 10^5)
6.01 × 10^5
= 601000 N
