With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s
Using the principles of free fall motion, it can be determined that a tennis ball dropped from a height of 1.16 meters would hit the ground with a velocity of 4.8 m/s.
The objective here is to calculate the initial velocity of the tennis ball just before it hits the ground. This is a physics question related to free fall motion. We will use the equation of motion, v^2 = u^2 + 2as, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration, and 's' is the distance. In our case, the initial velocity (u) is 0 because the ball is dropped, not thrown downwards.
The acceleration (a) is the acceleration due to gravity, which is -9.8 m/s² (it's negative because it acts downwards). The distance (s) will be the drop height, which is -1.16 m (it's negative because we're considering downwards as negative direction). Therefore, the equation becomes (v)^2 = 0 + 2*(-9.8 m/s²)*(-1.16 m). Solving this we get v = √(2*9.8*1.16) m/s = 4.8 m/s. So, the tennis ball hits the ground with a velocity of 4.8 m/s.
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There is a chemical change because the liquid water, after passing an electric current, produces hydrogen and oxygen gases. This is a reaction. The molecular component of eater has been broken down into two. These two gases have different structures compared to water.
a. True
b. False
b. False
Oersted
Faraday
Ampere