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3 years ago

9

Why won't Brainly let me make a account or log in? It's always telling me that it can't take my registration at this time or it

won't work when I've filled everything out

1 answer:

patriot [66]3 years ago

5 0

Answer:

I don't know. It does the same thing to me to. I had to create a new account with my friends email because it won't let me log in

Explanation:

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Whats the best used for Arch bridge

SOVA2 [1]

Answer:

China has also been constructing arch bridges for many years. They built the Zhaozhou Bridge in 605 AD, and it is still around today! Modern arch bridges use materials such as concrete, steel and iron.

Explanation:

6 0

3 years ago

What is the work required to deflect a linear spring (k=32 kN/m) by 120 cm?

Anit [1.1K]

Answer:

the work done by the linear spring will be equal to 23.04 k J

Explanation:

given,

k = 32 k N /m

deflected spring = 120 cm

= 1.2 m

work done by spring can be calculate as

= $\dfrac{1}{2}kx^2$

= $\dfrac{1}{2}\times 32 \times 1.2^2$

= 23.04 k J

hence, the work done by the linear spring will be equal to 23.04 k J

6 0

3 years ago

Hello it's my new id<br>I am numu

Sonja [21]

Answer:

i am felix

Exp lanation:

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6 0

3 years ago

a metal coin has certain properties that can be measured.which property of a coin is different on the moon that is on earth?

Sloan [31]

Answer:

Coins weigh less on the Moon.

Explanation:

Gravity is only 1/6th as strong on the Moon than it is on Earth. Where a nickle is about 5 grams on Earth, it is less than 1 gram on the Moon. Gravity is affected by the size of the planet or moon. The Moon is much less massive than the Earth.

8 0

3 years ago

A rotating cup viscometer has an inner cylinder diameter of 2.00 in., and the gap between cups is 0.2 in. The inner cylinder len

Lesechka [4]

Answer:

The dynamic viscosity and kinematic viscosity are $1.3374\times 10^{-6}$ lb-s/in2 and $1.4012\times 10^{-3}$ in2/s.

Explanation:

Step1

Given:

Inner diameter is 2.00 in.

Gap between cups is 0.2 in.

Length of the cylinder is 2.5 in.

Rotation of cylinder is 10 rev/min.

Torque is 0.00011 in-lbf.

Density of the fluid is 850 kg/m3 or 0.00095444 slog/in³.

Step2

Calculation:

Tangential force is calculated as follows:

T= Fr

$0.00011 = F\times(\frac{2}{2})$

F = 0.00011 lb.

Step3

Tangential velocity is calculated as follows:

$V=\omega r$

$V=(\frac{2\pi N}{60})r$

$V=(\frac{2\pi \times10}{60})\times1$

V=1.0472 in/s.

Step4

Apply Newton’s law of viscosity for dynamic viscosity as follows:

$F=\mu A\frac{V}{y}$

$F=\mu (\pi dl)\frac{V}{y}$

$0.00011=\mu (\pi\times2\times2.5)\frac{1.0472}{0.2}$

$\mu =1.3374\times 10^{-6}$ lb-s/in².

Step5

Kinematic viscosity is calculated as follows:

$\upsilon=\frac{\mu}{\rho}$

$\upsilon=\frac{1.3374\times 10^{-6}}{0.00095444}$

$\upsilon=1.4012\times 10^{-3}$ in2/s.

Thus, the dynamic viscosity and kinematic viscosity are $1.3374\times 10^{-6}$ lb-s/in2 and $1.4012\times 10^{-3}$ in2/s.

4 0

4 years ago