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Elis [28]
3 years ago
6

Use Poiseuille's Law to calculate the rate of flow in a small human artery where we can take η = 0.028, R = 0.008 cm, l = 2 cm,

and P = 5000 dynes/cm2. (Round your answer to three significant figures.)
Physics
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

1.435 x 10^-4 cm^3/s

Explanation:

coefficient of viscosity, η = 0.028 poise

Radius, R = 0.008 cm

length, l = 2 cm

Pressure, P = 5000 dyne/cm^2

The Poiseuille's formula is given below

V=\frac{\pi PR^{4}}{8\eta l}

Where, V is the rate of flow, that means volume flowing per second

V=\frac{3.14\times5000 \times {0.008}^{4}}{8 \times 0.028 \times 2}

V = 1.435 x 10^-4 cm^3/s

Thus, the rate of flow is 1.435 x 10^-4 cm^3/s.

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Two identical cars are parked in the sun on a warm day. One car is black and the other is white. Both cars are parked for three
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Answer:

black will be hotter because black paint absorbs light and white reflects it.

4 0
2 years ago
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A diamond with a mass of 45 g hangs motionless from a chain. what is the upward force of the chain on the diamond?
Oksi-84 [34.3K]
The upward force of the chain on the diamond would be the tension in the chain, and this tension would have to support the weight of the 45g that hangs from the chain.

mass = 45 g = 45/1000 kg = 0.045kg

Weight = mg = 0.045 * 10 ≈ 0.45N,            g ≈ 10 m/s²

<span>So the upward force is ≈ </span><span>0.45N. </span>
6 0
3 years ago
A 2,000 g quantity of C-14 is left to undergo radioactive decay. The half-life of C-14 is approximately 5,700 years. After going
Katyanochek1 [597]
Start with 2,000 grams.
After 1 half-life, 1,000 grams are left.
After another half-life, 500 are left.
After another half-life, 250 are left.
After another half-life, 125 are left.

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X = 4 .
8 0
3 years ago
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Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

  • V_r = resultant velocity of the plane
  • \theta = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

5 0
3 years ago
You can think of the work-kinetic energy theorem as the second theory of motion, parallel to Newton's laws in describing how out
kiruha [24]

Answer:

a) 4 289.8 J

b) 4 289.8 J

c) 6 620.1 N

d) 411 186.3 m/s^2

e) 6 620.1 N

Explanation:

Hi:

a)

The kinetic energy of the bullet is given by the following formula:

K = (1/2) m * v^2

With

    m = 16.1 g = 1.61 x 10^-2 kg

     v = 730 m/s

K = 4 289.8 J

b)

the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest)  Ki = 0, therefore:

W = ΔK = Kf - Ki  = 4 289.8 J

c)

The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).

If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:

W = F * L

Where F is the net force and L is the length of the barrel, that is:

F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N

d)

The acceleration can be found dividing the force by the mass:

a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2

e)

The force will have a magnitude equal to c) and direction along the barrel towards the exit

5 0
3 years ago
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