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Ugo [173]
3 years ago
8

Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).

Physics
1 answer:
Oliga [24]3 years ago
6 0

To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}

Where

P = Pressure at each point

r = Radius

\eta = Viscosity

l = Length

Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}

From the problem two terms are given

R_A = \frac{R_B}{2}

L_A = 2L_B

Replacing we have to

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

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f your risk-aversion coefficient is A = 4 and you believe that the entire 1926–2015 period is representative of future expected
siniylev [52]

Answer:

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

Explanation:

As the complete question is not given here ,the table of data is missing which is as attached herewith.

From the maximized equation of the utility function it is evident that

Weight=\frac{E_M-r_f}{A\sigma_M^2}

For the equity, here as

  • Weight is percentage of the equity which is to be calculated
  • {E_M-r_f} is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
  • A is the risk aversion factor which is given as 4.
  • \sigma_M is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59

By substituting values.

Weight=\frac{E_M-r_f}{A\sigma_M^2}\\Weight=\frac{8.30\%}{4(20.59\%)^2}\\Weight=0.4894 =48.94\%

So the weight of equity is 48.94%.

Now the weight of T bills is given as

Weight_{T-Bills}=1-Weight_{equity}\\Weight_{T-Bills}=1-0.4894\\Weight_{T-Bills}=0.5105=51.05\%\\

So  the weight of T-bills is 51.05%.

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

7 0
3 years ago
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it i
lianna [129]

Answer:

<em>A) the moment of inertia of the system decreases and the angular speed increases. </em>

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the  merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

I_{1} w_{1} = I_{2} w_{2}    ....1

where I_{1} and I_{2} are the initial and final moment of inertia respectively.

and w_{1} and w_{2} are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

I = mr^{2}    ....2

where m is the mass of the rotating body,

and r is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from I_{1} to I_{2} will cause the angular speed of the system to increase from w_{1} to w_{2} .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

7 0
3 years ago
A man pushes an 250-N crate at constant speed a distance of 30.0 m upward along a rough slope that makes an angle of 60° with th
mel-nik [20]

Answer:

6495.19 Joule

Explanation:

F = Weight of the crate = 250 N

d = Distance the cart is pushed = 30 m

θ = Angle of inclination = 60°

The weight of the crate will be resloved into two components

Fdsinθ and Fdcosθ

Work done by the force of gravity is

W = Fdsinθ

⇒W = 250×30×sin60

⇒W = 6495.19 Joule

∴ The work done by the force of gravity is 6495.19 Joule

8 0
3 years ago
An electron follows a helical path in a uniform magnetic field of magnitude 0.340 T. The pitch of the path is 6.00 µm, and the m
dedylja [7]

Answer:

The electron's speed is 34007.35 m/s

Explanation:

It is given that,

Magnetic field, B = 0.34 T

Magnetic force on the electron, F=1.85\times 10^{-15}\ N

The electron follows a helical path. We have to find the speed of an electron. The formula for magnetic force is given by :

F=B\times q\times v

q = charge on an electron, q=1.6\times 10^{-19}\ C

v = velocity of an electron

v=\dfrac{F}{Bq}

v=\dfrac{1.85\times 10^{-15}\ N}{0.34\ T\times 1.6\times 10^{-19}\ C}

v = 34007.35 m/s

Hence, this is the required solution.

4 0
3 years ago
In the pulley system shown below, a 360 N weight is slowly lifted. Assuming the system is 100% efficient and each pulley is weig
STALIN [3.7K]
Your answer is D. 361 N
8 0
3 years ago
Read 2 more answers
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