Answer:
a) The speed of the block immediately after the collision is .
b) The impulse exerted on the block is .
Explanation:
Hi
a) As this is a perfectly elastic collision, we can use the formula , due , we obtain . Then with the data that we know and , therefore or adding uncertainty.
b) Now that we know the speed we can use
Answer:
The horizontal force acting on m2 is F + 9.8m1
Explanation:
Given;
Block m1 on left of block m2
Make a sketch of this problem;
F →→→→→→→→→→→-------m1--------m2
Apply Newton's second law of motion;
F = ma
where;
m is the total mass of the body
a is the acceleration of the body
The horizontal force acting on block m2 is the force applied to block m1 and force due to weight of block m1
F₂ = F + W1
F₂ = F + m1g
F₂ = F + 9.8m1
Therefore, the horizontal force acting on m2 is F + 9.8m1
False because their are single cell organisms
Answer: AAAAAAAAGGGGGHHHHJJJGSSSUUUUUUUUYCCFVGBHNJM
Explanation: YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET
Answer:
Therefore, highest point that the cannon ball reaches is 168.7m
Explanation:
the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s
highest point that the cannon ball reaches?
g = 9.8m/s2
Therefore, highest point that the cannon ball reaches is 168.7m