This question deals with the law of conservation of momentum, which basically says that the total momentum in a system must stay the same, provided there are no outside forces. Since you were given the mass and velocity of the two objects you can find the momentum (p=mv) of each and then add them together to find the total momentum of the system before they collide. This total momentum must be the same after they collide. Since you have the mass and velocity of one of the objects after the collision you can find the its momentum after. Subtract this from the the system total and you will have the momentum of the other object after the collision. Now that you know the momentum of the other object you can find its velocity using p=mv and its mass from before.
Be careful with the velocities. They are vectors, so direction matters. Typically moving to the right is positive (+) and moving to the left is negative (-). It is not clear from your question which direction the objects are moving before and after the collision.
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:

Explanation:
It is given that,
Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.


Initial speed of the projectile is v and final speed is 0.5 v.


g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :



So, the height of the projectile above the ground is
. Hence, this is the required solution.
0.04m²
Explanation:
Given parameters:
Pressure = 250000Pa
Weight = 40000N
Unknown:
Area of each foot = ?
Solution:
Pressure is the force exerted per unit area of a body
Pressure = 
To find the area;
Area = 
Area =
= 0.16m²
The force exerted by all the four feet is 0.16m²
the area of each feet =
= 0.04m²
Learn more:
Pressure brainly.com/question/7139767
#learnwithBrainly
Missing figure of the problem: http://tsephysics.weebly.com/uploads/5/1/9/3/51934203/477140_orig.jpg
Solution:
Assuming the potential is zero at infinite distance from the charge, then the potential at a certain distance r from a single point charge is

where

is the Coulomb's constant.
In our problem, we just have to superimpose the potential generated by every charge. The diagonal of the square is

, therefore the distance between each charge and the center of the square is

.
So, the total potential is:

