Respuesta :
There is an omission of some sentences in the question which affects the answering of question B and C, so we will based the omission of the sentences on assumption in order to solve the question that falls under it.
NOTE: The omitted sentences are written in bold format
A tennis ball has a mass of 0.059 kg. A professional tennis player hits the ball hard enough to give it a speed of 41 m/s (about 92 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (41 m/s).
As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.0 cm at the instant when its speed is momentarily zero, before rebounding.
A) What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero?
B) How much time elapses between first contact with the wall, and coming to a stop?
C) What is the magnitude of the average force exerted by the wall on the bal dring contact?
D) In contrast, what is the magnitude of the gravitational force of the Earth on the ball?
Answer:
a) [tex]V_{avg} = 20.5m/s[/tex]
b) 9.76 × 10⁻⁴s
c) 247.9 N
d) 5.8 N
Explanation:
Given that;
Initial speed [tex](V_i)[/tex] = 0
Final speed [tex](V_f)[/tex] = 41 m/s
Distance (d) = 0.002
mass (m) = 0.059 kg
g = 9.8 m/s²
a)
The average speed of the ball can be calculated as;
[tex]V_{avg} = \frac{V_i+V_f}{2}[/tex]
[tex]V_{avg} = \frac{0+41}{2}[/tex]
[tex]V_{avg} = 20.5m/s[/tex]
b)
The time elapsed can be calculated by using the second equation of motion which is given as:
[tex]S=(\frac{V_i+V_f}{2})t[/tex]
If we make time (t) the subject of the formula; we have:
[tex](V_i+V_f)t=2S[/tex]
[tex]t= (\frac{2S}{V_I+V_f})[/tex]
[tex]=\frac{2(0.02)}{41+0}[/tex]
[tex]=\frac{0.04}{41}[/tex]
= 0.000976
=9.76 × 10⁻⁴s
c)
the magnitude of the average force (F) exerted by the wall on the bal dring contact can be determined using;
Force (F) = mass × acceleration
where acceleration [tex](a)= \frac{Vo}{t}[/tex]
[tex]\frac{41}{0.00976}[/tex]
acceleration (a) = 4200.82 m/s²
F = m × a
= 0.059 × 4200.82
= 247.85
≅ 247.9 N
d)
the magnitude of the gravitational force of the Earth on the ball
Force (F) = mass (m) × gravity (g)
= 0.059kg × 9.8 m/s²
= 5.782 N
≅ 5.8 N
The average speed of the ball is 41 m/s.
The time of motion of the ball is 9.75 x 10⁻⁴ s.
The magnitude of the average force exerted by the wall on the ball during contact is 2,481.03 N.
The magnitude of the gravitational force of the earth on the ball is 0.58 N.
The given parameters:
- Mass of the tennis ball, m = 0.059 kg
- Speed of the ball, u = 41 m/s, left
- Final speed of the ball, v = 41 m/s, right
The average speed of the ball is calculated as follows;
[tex]v = 41 \ m/s - 0\\\\ v = 41 \ m/s[/tex]
The time of motion of the ball is calculated as follows;
[tex]s = (\frac{v_f + v_0}{2} )t\\\\ 0.02 = (\frac{0 + 41}{2} )t\\\\ t = \frac{2(0.02)}{41} \\\\ t = 9.75 \times 10^{-4} \ s[/tex]
The magnitude of the average force exerted by the wall on the ball during contact is calculated as follows;
[tex]F = \frac{m \Delta v}{t} \\\\ F = \frac{0.059 \times 41}{9.75 \times 10^{-4} } \\\\ F = 2,481.03 \ N[/tex]
The magnitude of the gravitational force of the earth on the ball is calculated as follows;
[tex]F = mg\\\\ F = 0.059 \times 9.8\\\\ F = 0.58 \ N[/tex]
The missing part of the question is below:
As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.0 cm at the instant when its speed is momentarily zero, before rebounding.
Learn more about average force and speed here: https://brainly.com/question/6504879
