Answer:
The velocity after 2 seconds can be found through:
V = u +a*t
Where V is final velocity, u is initial velocity, a is acceleration and t is time.
V = 0 + 2* 2= 4 meters/second
The distance (s) can be found through:
V^2= u^2 +2*a* s
Where V is final velocity, u is initial velocity, a is acceleration.
4^2= 0^2 + 2 *2*s
16= 0 + 4s
s= 4 meters
Distance (s) can also be found through:
s= ut + 1/2 at^2
s= 0+ 1/2 *2*2^2= 1 *2*2
s= 4 meters
Explanation:
Answer:
Direction of ship: 9.45° West of North
Ship's relative speed: 7.87m/s
Explanation:
A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0
Vx=0
Therefore, -VsSin∅+VcCos∅40°
Sin∅ = Vc/Vs × Cos 40°
Sin∅ = 1.5/7 ×Cos40°
Sin∅= 0.164
∅= Sin-¹ (0.164)
∅= 9.45° W of N
B. Ship's relative speed:
Vy= VsCos∅ + Vcsin40°
= 7Cos9.45° + 1.5sin40°
= 7×0.986 + 1.5×0.642
= 7.865
= 7.87m/s
Answer:
The torque about the origin is 
Explanation:
Torque
is the cross product between force
and vector position
respect a fixed point (in our case the origin):

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:
![\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]](https://tex.z-dn.net/?f=%20%5Coverrightarrow%7Br%7D%5Ctimes%5Coverrightarrow%7BF%7D%20%3D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Chat%7Bi%7D%20%26%20%5Chat%7Bj%7D%20%26%20%5Chat%7Bk%7D%5C%5C%20F1_%7Bx%7D%20%26%20F1_%7By%7D%20%26%20F1_%7Bz%7D%5C%5C%20r_%7Bx%7D%20%26%20r_%7By%7D%20%26%20r_%7Bz%7D%5Cend%7Barray%7D%5Cright%5D%20)



unlock your third eye and youll be able to see it
Answer:
8.27°
Explanation:
To angle difference will be determined by the difference in the displacement of the springs, produced by the weight of the center of mass of the rod.
![d=y_1-y_2=\frac{F_1}{k_1}-\frac{F_2}{k_2}=\frac{0.5mg}{31N/m}-\frac{0.5mg}{63N/m}\\\\d=0.5(1.6kg)(9.8m/s^2)[\frac{1}{31N/m}-\frac{1}{63N/m}]=0.128m](https://tex.z-dn.net/?f=d%3Dy_1-y_2%3D%5Cfrac%7BF_1%7D%7Bk_1%7D-%5Cfrac%7BF_2%7D%7Bk_2%7D%3D%5Cfrac%7B0.5mg%7D%7B31N%2Fm%7D-%5Cfrac%7B0.5mg%7D%7B63N%2Fm%7D%5C%5C%5C%5Cd%3D0.5%281.6kg%29%289.8m%2Fs%5E2%29%5B%5Cfrac%7B1%7D%7B31N%2Fm%7D-%5Cfrac%7B1%7D%7B63N%2Fm%7D%5D%3D0.128m)
by a simple trigonometric relation you obtain that the angle:

hence, the angle between the rod and the horizontal is 8.27°