Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s
We have to find final velocity.
The equation we use is
Final velocity=initial velocity+acceleration x time
Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s
We would round this to:
Vf (final velocity)=2.7m/s
Answer:
Pressure applied to the needle is 7528 Pa
Explanation:
As we know by poiseuille's law of flow of liquid through a cylindrical pipe
the rate of flow through the pipe is given as

now we know that

radius = 0.2 mm
Length = 6.32 cm

now we have



now we have


Answer:
the cell is the smallest unit
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m