Answer:
The diameter of the needle is <u>4.675 cm</u>.
Explanation:
Given:
Volume flow rate is, 
Velocity of air expelled by pump is, 
Let the area of the needle be 'A' cm² and the diameter be 'd' cm.
We know that, volume flow rate of the air expelled by pump is given as the product of the needle's area and velocity of air flowing through that area.
Therefore, volume flow rate is given as:

Now, considering the needle to be circular, area of the needle can be written as:

Therefore, the diameter of the needle is 4.675 cm.
Answer:
76.73 ft/s
Explanation:
Let the final velocity is v.
initial velocity, u = 96 ft/s
g = 32 ft/s²
height, h = 52 feet
use third equation of motion
v² = u² - 2 gh
v² = 96 x 96 - 2 x 32 x 52
v = 76.73 ft/s
Thus, the speed of the ball as it reaches the ground is 76.73 ft/s.
Answer:
The height of the building is approximately 156.58 m
Explanation:
The mass of the ball dropped from rest from the building top = 0.660 kg
The time in which the ball falls, t = 5.65 seconds
The height, h, of the building is given from the following equation of motion;
h = u·t + ¹/₂·g·t²
Where;
u = The initial velocity of the ball = 0 m/s
g = The acceleration due to gravity = 9.81 m/s²
Plugging in the values, we have;
h = 0 × 5.65 + ¹/₂ × 9.81 × 5.65² ≈ 156.58 m
The height of the building, h ≈ 156.58 m.
I believe it is respiration