A tennis player hits a 0.1-kilogram tennis ball with her racket from the north end of a court. Her racket was traveling at 65 kilometers per hour. The ball accelerated at a rate of 10 meters per second squared. The ball hits the floor on the south end of the tennis court. The floor sends the ball toward the tennis player's opponent with the same acceleration. How much force did the floor on the south end of the court exert on the ball?(1 point)
A. 1N
B. 7.5N
C. 0.01N
D. 6.5N
The answers are
1 newton
2 force = mass × acceleration
3 1.6N
4 1n
5 8n
unknown is 100% correct!!! I was worried for a second .-.
Connexus kids uniting over here lol
Also Unknown is 100% correct
Thanks Man/Sis/Sib
unknown is right
To solve this question, we need to use Newton's second law of motion, which states that force (F) equals mass (m) multiplied by acceleration (a). In this case, we need to find the force exerted by the floor on the ball.
First, let's find the initial velocity of the ball. We are given that the player's racket was traveling at 65 kilometers per hour, which is equivalent to 18.06 meters per second (since 1 kilometer per hour is equal to 0.2778 meters per second).
Now, let's calculate the time it takes for the ball to hit the floor on the south end of the court. Since we know the ball's initial velocity and acceleration, we can use the kinematic equation: final velocity (vf) equals initial velocity (vi) plus acceleration (a) multiplied by time (t). In this case, the final velocity is 0 (since the ball hits the floor and stops moving vertically), so we can rearrange the equation to solve for time (t).
0 = 18.06 + (-10)t
-18.06 = -10t
t = 1.806 seconds
Next, we can use the kinematic equation: displacement (d) equals initial velocity (vi) multiplied by time (t) plus one-half multiplied by acceleration (a) multiplied by time (t) squared. Since the ball starts from the north end of the court and ends up on the south end, the displacement is the length of the tennis court, which we'll assume is 23.78 meters (standard size for a tennis court).
23.78 = 18.06t + (1/2)(-10)(t^2)
23.78 = 36.468t - 5t^2
5t^2 - 36.468t + 23.78 = 0
We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 5, b = -36.468, and c = 23.78. Plugging these values into the quadratic formula, we find:
t ≈ 0.727 seconds or t ≈ 5.193 seconds
Since t cannot be negative and the ball traveled from the north to the south end of the court, we'll use t = 5.193 seconds.
Now, we can find the force exerted by the floor on the ball. From Newton's second law of motion, we know:
F = m × a
The mass of the ball is given as 0.1 kilograms, and the acceleration is -10 meters per second squared (negative because the ball is moving in the opposite direction). Substituting these values into the formula, we get:
F = 0.1 × (-10)
F = -1N
Thus, the force exerted by the floor on the ball is -1N. However, since the question asks for the magnitude of the force (positive value), we take the absolute value of -1N, which is 1N.
Therefore, the correct answer is A. 1N.