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 player on the south end of the court hit the tennis ball with her racket at 75 kilometers per hour. How much force did the racket on the south end of the court exert on the ball?(1 point) A.1 N B.6.5 N C.7.5 N D.0.01 N

Explanation:

Given that:

Mass of ball = 0.1-kilogram

From the North End:

Acceleration of racket = 65 km/hour

Converting to m/seconds

= 65×1000/3600

65000/3600

= 18.055m/s

Ball acceleration= 10 meters per second squared

From the South End:

Acceleration of racket = 75 km/hour

Converting to m/seconds

75×1000/3600

= 75000/3600

= 20.833m/s

Now,

18.055m/s = 10m/s^2

20.833m/s = x

18.055m/s × x = 20.833m/s × 10m/s^2

18.055m/s × x = 208.33

x = 208.33/18.055m/s

X = 11.538m/s^2

Hence, the ball accelerated at the rate of 11.538m/s^2 in the South End.

F = ma

Where a = 11.538m/s^2

m = 0.1kg

F = 0.1×11.538m/s^2

F = 1.15N

f is prop to m1xm2 double m1 double force

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.