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Gala2k [10]
3 years ago
7

The radius of the aorta is about 1 cm and the blood flowing through it has a speed of about 30 !" ! . Calculate the average spee

d of the blood in the capillaries given the total cross section of all the capillaries is about 2000 !"!
Physics
1 answer:
Julli [10]3 years ago
4 0

Answer:

The average speed of the blood in the capillaries is 0.047 cm/s.

Note: The question is incomplete. The complete question is as follows:

The radius of the aorta is about 1 cm and the blood flowing through it has a speed of about 30 cm/s. Calculate the average speed of the blood in the capillaries given the total cross section of all the capillaries is about 2000 cm².

Explanation:

From the given values:

radius of the aorta, r₁ = 1 cm

speed of blood, v₁ = 30 cm/s

Area of the aorta, A₁ = πr₁² where π = 3.142

Area of aorta = 3.142 × (1)² = 3.142 cm²

Area of the capillaries, A₂ = 2000 cm²

let the average speed of the blood in the capillaries = v₂

From the continuity equation of fluid flow, the product of cross-sectional area of the pipe and the fluid speed at any point along the pipe is always constant. In formula, A₁v₁ = A₂v₂

Using the continuity equation, the average the average speed of the blood in the capillaries can be calculated thus:

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.

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A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.
Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

         = 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = 0.5mv^{2}

           = 0.5 x 25 x 5.6^{2}  = 392 J

(C) average frictional force = \frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}

  • change in KE (ΔKE) = initial KE - final KE
  • ΔKE = 0.5mv^{2} - 0.5mvf^{2}            
  • when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes  ΔKE = 0.5x25x5.6^{2} - 0 = 392 J

 \frac{-(ΔKE+ΔU)}{h}[/tex] = \frac{-(392 + (-2940))}{12}

=  \frac{(-392 + 2940)}{12} = 212.33 N

5 0
3 years ago
.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric
maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0
3 years ago
the moon revolves around the earth in a nearly circular orbit kept by gravitational force exerted by the earth work done will be
rodikova [14]

Answer:

Zero because the applied force is perpendicular to the motion of the object.

No work is done on an object moving is a circular path about a central attractive force.

Any work done in such a case would result in a change in the orbit.

3 0
2 years ago
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Serjik [45]
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