The velocity of tennis racket after collision is 14.96m/s
<u>Explanation:</u>
Given-
Mass, m = 0.311kg
u1 = 30.3m/s
m2 = 0.057kg
u2 = 19.2m/s
Since m2 is moving in opposite direction, u2 = -19.2m/s
Velocity of m1 after collision = ?
Let the velocity of m1 after collision be v
After collision the momentum is conserved.
Therefore,
m1u1 - m2u2 = m1v1 + m2v2
Therefore, the velocity of tennis racket after collision is 14.96m/s
Answer:
Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc) items 3 and 4 are eliminated.
Again item 2 refers to bonds between atoms and is eliminated.
This leaves only item 1.
Nuclear forces are very short range forces between components of the nucleus.
Weak nuclear forces are trillions of times smaller than strong forces.
Gravitational forces are much much smaller than the weak nuclear force.
A bicyclist can ride their bicycle still on the road. Bicycle riders be able to take the public ways which has the similar rights and accountability as motorists and are subject to the same guidelines and protocols. The law says that individuals who ride bikes should ride as nearby to the right side of the road as likely excluding under the following conditions: when passing, preparing for a left go, evading risks, if the lane is too constricted to share, or if oncoming a place where a right turn is approved. In a road which has a bike lane the bicyclists roving slower than road traffic must custom the bike way excluding when creating a left turn, passing, evading hazardous settings, or impending a place where a right turn is approved.
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.