The velocity of the submarine immediately after firing the missile is 0.0104 m/s
Explanation:
Mass of the submarine M=50 tonne=
Mass of the missile m=40 kg
velocity of the missile v= 13m/s
we have to calculate the velocity of the submarine after firing
This is the recoil velocity and its expression is derived from the law of conservation of momentum
recoil velocity of the submarine

Answer:
3: I can´t see the text/image, but it depend on the mass and the force applied to the ball, if both are too high, it will be harder to make a home run. (Second law)
4:It would be easier to make a home run because there is no interruption between the ball and the space the same travels. (Third law)
Explanation:
Answer:
6.746 ft/s^2
Explanation:
v(t)=50
v(0)=27
t=5/3600 = 1/720 hours
v(t)-v(0)= a(t-0)
50-27= a(1/720)
a= 23*720= 16560 mi/h^2
16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2
Assuming this is an elastic collision, they go in the direction of the object with more mass
One-dimensional motion can be plotted through the Cartesian plane which has a coordinates of (x,y). These coordinates are the abscissa and ordinates. Since, there are two coordinates, the answer to the second item is two.
The symbol that can be used to identify systems position is (x,y). Since this is one dimensional motion, it is possible that one of the two coordinates becomes zero.