Where are the following sketches?
Answer:
The moment (torque) is given by the following equation:
![\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)](https://tex.z-dn.net/?f=%5Cvec%7B%5Ctau%7D%20%3D%20%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7BF%7D%5C%5C%5Cvec%7Br%7D%20%5Ctimes%20%5Cvec%7BF%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5C%5E%7Bi%7D%26%5C%5Ej%26%5C%5Ek%5C%5Cr_x%26r_y%26r_z%5C%5CF_x%26F_y%26F_z%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5C%5E%7Bi%7D%26%5C%5Ej%26%5C3k%5C%5C0.23%260.04%260%5C%5C150%26260%260%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5C%5Ek%28%280.23%2A260%29%20-%20%280.04%2A150%29%29%20%3D%20%5C%5Ek%20%2853.8~Nm%29)
Explanation:
The cross-product between the distance and the force can be calculated using the method of determinant. Since the z-components are zero, it is easy to calculate.
Answer:
vB' = 0.075[m/s]
Explanation:
We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.
Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

where:
mA = 0.355 [kg]
vA = 0.095 [m/s] before the collision
mB = 0.710 [kg]
vB = 0.045 [m/s] before the collision
vA' = 0.035 [m/s] after the collision
vB' [m/s] after the collison.
The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.
![(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]](https://tex.z-dn.net/?f=%280.355%2A0.095%29%2B%280.710%2A0.045%29%3D%280.355%2A0.035%29%2B%280.710%2Av_%7BB%27%7D%29%5C%5Cv_%7BB%27%7D%3D0.075%5Bm%2Fs%5D)
Answer:
v = √2G
/ R
Explanation:
For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)
Eo = K + U = ½ m1 v² - G m1 m2 / r1
Ef = - G m1 m2 / r2
When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf
Eo = Ef
½ m1v² - G m1
/ R = - G m1
/ R
v² = 2G
(1 / R - 1 / Rinf)
If we do Rinf = infinity 1 / Rinf = 0
v = √2G
/ R
Ef = = - G m1 m2 / R
The mechanical energy is conserved
Em = -G m1
/ R
Em = - G m1
/ R
R = int ⇒ Em = 0