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
Answer:
B) 100 J
Explanation:
Assuming the distance given is measured along the incline, the vertical change in height is (5 m)(sin 30°) = 2.5 m. Then the change in potential energy is ...
∆PE = mg(∆h) = (4 kg)(10 m/s^2)(2.5 m) = 100 J
<h3><u>Volume is 0.1848 m³</u></h3><h3 />
Explanation:
<h2>Given:</h2>
m = 49.9 kg
ρ = 270 kg/m³
<h2>Required:</h2>
volume
<h2>Equation:</h2>
where: ρ - density
m - mass
v - volume
<h2>Solution:</h2>
Substitute the value of ρ and m
<h2>Final Answer:</h2><h3><u>Volume is 0.1848 m³</u></h3>
The question seems to be what is an equilibrant force.
The answer is "an added force that produces equilibrium.
Here you have more insight:
<span>an object that has no net force acting on it? This object indeed is in equilibrium but the object is not the equilibran force.
the reaction force in an action-reaction pair of forces?
the reaction force is not an equilibrant force. The reaction force exists always but equilibrium is only possible if the net force is cero.
an added force that produces equilibrium? this is the right answer.</span>
