All categories

3 years ago

A liquid's viscosity increases with increasing temperature. a)-True b)-False

Engineering

1 answer:

olasank [31]3 years ago

7 0

Answer:

b) False

Explanation:

Viscosity:

Viscosity is a fluid property and comes in the picture when fluid in the motion.In Simple words viscosity is the frictional force offered by fluid between the fluid layer.Viscosity provides a resistant to flow of fluid.

Generally viscosity are of two types

1.Dynamics viscosity

2.Kinematics viscosity

Generally in liquids when temperature of fluid is increases then molecular force between fluid particle goes to decreases.Due to this viscosity of liquids will decrease.

So our option b is right.

You might be interested in

Hello , how are yall:))))

SVEN [57.7K]

Answer:

eh I'm good hbu?????????

6 0

3 years ago

Read 2 more answers

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at (100%) (125

Anna [14]

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.

Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:

Without any extra adjustments or corrections, either 125% of the continuous load, OR

When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).

This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.

To know more about connectors click here:

brainly.com/question/16987039

#SPJ4

4 0

1 year ago

A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).

Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress $\sigma_m = 90MPa$

stress amplitude $\sigma_a = 190MPa$

a) $\sigma_m =\frac{\sigma_max+\sigma_min}{2}$

$90 =\frac{\sigma_{max}+\sigma_{min}}{2}$ --------------1

$\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}$

$190 = \frac{\sigma_{max}-\sigma_{min}}{2}$ -----------2

solving 1 and 2 equation we get

$\sigma_{max} = 280MPa$

b) $\sigma_{min} = - 100MPa$

stress ratio $=\frac{\sigma_{min}}{\sigma_{max}}$

$=\frac{-100}{280} = -0.35$

d)magnitude of stress range

$=(\sigma_{max} -\sigma_{min})$

= 280 -(-100) = 380 MPa

3 0

3 years ago

Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water

olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate = 2 L/s

water temperature 30 degree celcius

$Q = A\times V$

$2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity$

$V = \frac{20\times 4}{9\times \pi} = 2.83 m/s$

$Re = \frac{\rho\times V\times D}{\mu}$

at 30 degree celcius $= \mu = 0.798\times 10^{-3}Pa-s , K = 0.6154$

$Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}$

Re = 106390

So ,this is turbulent flow

$Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}$

$Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419$

$\frac{h\times 0.03}{0.615} = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}$

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 0

3 years ago

A man can swim at 4 ft/s in still water. He wishes to cross tje 40-ft wide river to point B, 30 ft downstream. If the river flow

iVinArrow [24]

Answer:

$v \approx 4.472\,\frac{ft}{s}$ , $t = 10\,s$

Explanation:

Since man and river report constant speeds and velocities are mutually perpendicullar, the absolute speed of the man is calculated by the Pythagorean Theorem:

$v = \sqrt{(4\,\frac{ft}{s} )^{2}+(2\,\frac{ft}{s} )^{2}}$

$v \approx 4.472\,\frac{ft}{s}$

The required time to make the crossing is:

$t = \frac{40\,ft}{4\,\frac{ft}{s} }$

$t = 10\,s$

6 0

3 years ago