Answer:
The average speed of the blood in the capillaries is 0.047 cm/s.
Explanation:
Given;
radius of the aorta, r₁ = 1 cm
speed of blood, v₁ = 30 cm/s
Area of the aorta, A₁ = πr₁² = π(1)² = 3.142 cm²
Area of the capillaries, A₂ = 2000 cm²
let the average speed of the blood in the capillaries = v₂
Apply continuity equation to determine the average speed of the blood in the capillaries.
A₁v₁ = A₂v₂
v₂ = (A₁v₁) / (A₂)
v₂ = (3.142 x 30) / (2000)
v₂ = 0.047 cm/s
Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.
The slope of a speed-time graph is the acceleration represented by the graph.
All other parts of this question refer to a lab experiment or exercise
where I was not present, but Zeesam16 was. Therefore I have no data
with which to answer the rest of the question, and hope that Zeesam can
handle it.
friction is the resistance that one surface or object encounters when moving over another. Due to gravity pulling everything down things need to friction in order to move
i hope this helps :/

GiveN:
- Initial velocity = 9.8 m/s²
- Accleration due to gravity = -9.8 m/s²
- Time taken = 1 s
To FinD:
- Final velocity of the ball?
Step-by-step Explanation:
Using the first Equation of motion,
⇒ v = u + gt
⇒ v = 9.8 + -9.8(1)
⇒ v = 0 m/s
The final velocity is hence <u>0</u><u> </u><u>m</u><u>/</u><u>s</u><u>.</u>
<h3>
Note:</h3>
- While solving questions of under gravity motions using equations of motion, remember the sign convection to avoid mistakes.
- You can consider positive above the ground and negative for towards it.
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m
K.E = 1/2 x 0.17 x 
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/
To calculate the final velocity, let us use third equation of motion.
=
+ 2as
=
+ 2 x 0.3 x 61
= 36 + 36
= 72
V = 
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: brainly.com/question/402617