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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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2 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
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Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
2 years ago
What is the sharpness of the light and the amount of voltage when you put the magnet through the lower number of coils first vs
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4 0
3 years ago
When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1lbm=0.453
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Answer:

a) 0.022%

b) 10014.32 lb

Explanation:

a) Percentage uncertainty would be

0.0001\times \frac{100}{0.4539}=0.022%

Percent uncertainty is 0.022%

b) For 1 kg uncertainty mass in kg would be

\frac{1}{0.022}\times {100}=4545.5\ kg

Mass in pounds would be

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8 0
3 years ago
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Answer:

Current will decrease.

Explanation:

When we increase the number of stepping in transformer, the voltage will increase as its is directly proportional to the number of turn of stepping. Thus as the voltage will increase, current will decrease. As per the equation of ideal transformer,   E1 / E2 = I2 / I1

E1 and E2 are the voltages in primary and secondary winding and I1 and I2 are the current.

As the number of turns will be increased more inevitable losses will be generated that dissipates heat thus warming the primary.

Though the conservation of energy is obeyed but losses occur in this scenario hence step-up transformers cannot be used to create free energy.

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