7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
when we find the distance we will add all the blocks so
distance = 6+6+4
distance = 14blocks
when we find the displacement we will add and minus too
As you can read he goes to the south 6 and to north 6 so he leave that place and back to the place again so the displacement is 0. and again he goes to the west 4 blocks so the displacement = <em><u>4blocks</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>west</u></em>
Answer:
Light is shown through air on a diamond n=2.42) and it partially reflects and reflects.
Based on the incident angle of 62.5 shown what is the angle of reflection of the light
question of options
21.5
Dinosaurs but I need the whole groups yo tell you ;)
Answer:
C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of Fr.
Explanation:
Yo want to prove the following equation:
That is, the net force exerted on an object is equal to the change in the kinetic energy of the object.
The previous equation is also equal to:
(1)
m: mass of the block
vf: final velocity
v_o: initial velocity
Ff: friction force
F(x): Force
x: distance
You know the values of vf, m and x.
In order to prove the equation (1) it is necessary that you have C The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F. Thus you can calculate experimentally both sides of the equation.
