Essay about Jean
1. How hard must you pull on the third rope to keep the knot from moving?
Put the knot at 0,0 and the 3 unit force along the x axis. The 5 unit force goes into the second quadrant
3 units * cos(ᶱ) = + 3 units
5 units * cos(120o) = -2.5 units.
Resultant = +3 - 2.5 = 0.5 units.
Sum the forces vertically
3*sin(ᶱ) = 0
5*sin(120o) = 4.33 units
Resultant + 4.33 vertically
Find where you should be.
F = sqrt(Fx^2 + Fy^2)
F = sqrt(4.33^2 + 0.5^2)
F = sqrt(19)
F = 4.35 N
So you must pull with 4.35 units of force.
Now for the angle the angle = 180 + Tan-1(vertical/horizontal) = 180 + tan-1(4.33/0.5) = 263.4 o from the + x axis. …show more content…
m=0.2 kg then the acceleration(a) a=Fm=F0.2 V=u+at from rest u=0 v=axt=F0.2x1 is the velocity after 1s p=mv=0.2xF0.2x1=Fkgm/s momentum mass m'=20k a'=Fm'=F20 V=u+at v'=a'xt=F20x1 so p'=m'xv'=20xF20x1=Fkgm/s both equal
7. How fast, in rpm, would a 100 g, 50-cm diameter beach ball have to spin to have an angular momentum of 0.10 kg m2/s?
The angular momentum of a particle of mass m with respect to a chosen origin is given by
L = mvr sin θ
0.10 = 0.1*v*0.025 v = 0.1 / 0.0025 v=40m/s=1578 rpm
8. You need to determine the density of a ceramic statue. If you suspend the statue from a spring scale, the scale reads 28.4 N. If you then completely submerge the statue in a tub of water, the scale reads 17.0 N. What is the statue’s density?
The statue is denser than water so it displaces an amount of water equal to its volume. and its weight on a spring scale is equal to its weight LESS the weight of the water it has displaced.
So, we have 28.4 - 17.0 = 11.4N of water displaced.
Now what is the volume of 11.4 N weight of water?
11.4/9.81 = 1.16 kg of water.
Now water has a density of 1000kg/m^3, so 1.16 kg= 1.16 x10^-3 m^3