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scoundrel [369]
3 years ago
15

Given that a fluid at 260°F has a kinematic viscosity of 145 mm^2/s, determine its kinematic viscosity in SUS at 260°F.

Physics
1 answer:
Oduvanchick [21]3 years ago
5 0

Answer:

kinematic viscosity in SUS is = 671.64 SUS

Explanation:

given data

kinetic viscosity = 145 mm^2/s

we know

1 mm = 0.1 cm

so kinetic viscosity in cm is \nu =145 (0.1)^{2} =1.45 cm^{2}/s

other unit of kinetic viscosity is centistokes

1 cm^{2}/s = 100 cst

so 1.45 cm^2/s will be 145 cst

if the temperature is 260°f , then cst value should be multiplied by 4.632. therefore kinematic viscosity in SUS is = 4.362 *145 = 671.64 SUS

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A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.
Kazeer [188]

Answer:

(a) Magnetic moment will be 17.212\times 10^{-4}A-m^2

(b) Torque will be 6.024\times 10^{-4}N-m

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2

Current is given as i=15mA=15\times 10^{-3}A

Number of turns N = 25

(A) We know that magnetic moment is given by magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by \tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m

4 0
3 years ago
Increase the slit width to 1050 nm. What happens to the width of the central bright fringe? (Always remember to wait a few secon
djyliett [7]

Answer:

b.

Explanation:

In case of Single Slit, diffraction will occur.

Then In Single slit Diffraction, width of central fringe is

x_c= 2D\lambda/a


where D= distance b/w screen and slit

a= slit width

\lambda = wavelength

Thus if Screen width increases keeping other factors same then width of central fringe becomes narrower as

x_c\propto 1/a

On increasing the slit width the central bright fringe width The width of the central bright fringe becomes narrower.

3 0
4 years ago
In our Solar System, the inner planets are rocky because Choose one: A. warm temperatures in the inner disk caused the inner pla
xeze [42]

The inner planets are rocky because The warm temperatures in the inner disk caused the inner planetesimals to be formed of mostly rocky material.

What are rocky planets?

  • Rocky planets are the planet in which constituents are mostly silicate rocks or metal. They are also regarded as a planet with a solid surface.
  • The formation of rocky planets is said to have occurred billions of years ago and its process of formation is termed accretion. Through accretion are its constituents formed as the more it goes bigger, the higher the rising temperature and pressure in its core and the elements which have to undergo accreted heat up, melt, and spread. Through this process, heavier elements go deeper into the core of the planet and lighter elements float toward the surface.
  • In the formation of rocky planets, the inner portions of the disk are said to be warm from the protostar thereby resulting in the production of the heavy elements that stay there.
  • Examples of rocky planets are Earth or Mars

Hence, from the above, we can say that,

The warm temperatures in the inner disk caused the inner planetesimals to be formed of mostly rocky material.

Here,

Option A is correct.

Learn more about rocky planets here:

<u>brainly.com/question/22392798</u>

#SPJ4

3 0
2 years ago
During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca
Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

a=\frac{v^2}{r}

r is the radius of the loop

on substituting the respective values, we get

8\times9.81=\frac{340^2}{r}

or

r = 1472.98 m

5 0
4 years ago
A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet
puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

3 0
4 years ago
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