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.
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
Gravity increased the downward speed (or decreases the upward speed) by 9.8 m/s every second.
21.2/9.8 = 2.2 seconds
Answer:
400m
Explanation:
Brainliest? :))
Let your initial displacement from your home to the store be
Dd
>
1 and your displacement from the store to your friend’s house
be Dd
>
2.
Given: Dd
>
1 = 200 m [N]; Dd
>
2 = 600 m [S]
Required: Dd
>
T
Analysis: Dd
>
T 5 Dd
>
1 1 Dd
>
2
Solution: Figure 6 shows the given vectors, with the tip of Dd
>
1
joined to the tail of Dd
>
2. The resultant vector Dd
>
T is drawn in red,
from the tail of Dd
>
1 to the tip of Dd
>
2. The direction of Dd
>
T is [S].
Dd
>
T measures 4 cm in length in Figure 6, so using the scale of
1 cm : 100 m, the actual magnitude of Dd
>
T is 400 m.
Statement: Relative to your starting point at your home, your
total displacement is 400 m [S].
Answer:
longitudinal and transverse.
Explanation:
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