Answer:
7kgm/s
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Let P1A and P1B be the initial momentum of the bodies A and B respectively
Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.
Based on the law:
P1A+P2A = P1B + P2B
Given P1A = 5kgm/s
P2A = 0kgm/s(ball B at rest before collision)
P2A = -2.0kgm/s (negative because it moves in the negative x direction)
P2B = ?
Substituting the values in the equation gives;
5+0 = -2+P2B
5+2 = P2B
P2B = 7kgm/s
Answer:
Object 2, which has a density of 1.9 g/cm3, since it has more density than freshwater.
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
