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UNO [17]
3 years ago
14

What type of oil pressure gauge should be used when

Engineering
1 answer:
liq [111]3 years ago
6 0

Answer:

The mechanical gauge would be the one for the job

Explanation:

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Please help me with this, picture.
Alenkasestr [34]
Maybe try 086 degrees
3 0
2 years ago
Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)
ArbitrLikvidat [17]

Answer:

\mathbf{P_x =25 \ watts}

\mathbf{x_{rmx} = 5 \ unit}

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) +  sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal  Asin (ωt), Power can be expressed as:

P= \dfrac{A^2}{2}

For the number of sinosoidial signals;

Power can be expressed as:

P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...

As such,

For x(t), Power  P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}

P_x = \dfrac{25}{2}+ \dfrac{25}{2}

P_x = \dfrac{50}{2}

\mathbf{P_x =25 \ watts}

For the number of sinosoidial signals;

RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...

For x(t), the RMS value is as follows:

x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }

x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }

x_{rmx }=\sqrt{(\dfrac{50}{2} )}

x_{rmx} =\sqrt{25}

\mathbf{x_{rmx} = 5 \ unit}

8 0
3 years ago
1kg of air (R 287 J/kgK) fills a weighted piston-cylinder device at 50kPa and 100°C. The device is cooled until the temperature
Reika [66]

Answer:

the work done during this cooling is −28.7 kJ

Explanation:

Given data

mass (m) = 1 kg

r = 287 J/kg-K

pressure ( p) = 50 kPa

temperature (T) = 100°C = ( 100 +273 ) = 373 K

to find out

the work done during this cooling

Solution

we know the first law of thermodynamics

pv = mRT     ....................1

here put value of p, m R and T and get volume v(a) when it initial stage in equation 1

50 v(a) = 1 × 0.287  × 373

v(a) = 107.051 / 50

v(a) = 2.1410 m³    .......................2

now we find out volume when temperature is  0°C

so put  put value of p, m R and T and get volume v(b) when temperature is cooled in equation 1

50 v(b) = 1 × 0.287  × 273

v(a) = 78.351 / 50

v(a) = 1.5670 m³    .......................3

by equation 2 and 3 we find out work done to integrate the p with respect to v i.e.

work done = \int\limits^a_b {p} \, dv

integrate it and we get

work done = p ( v(b) - v(a)  ) ................4

put the value p and v(a) and v(b) in equation 4 and we get

work done = p ( v(b) - v(a)  )

work done = 50 ( 1.5670 - 2.1410 )

work done = 50 ( 1.5670 - 2.1410 )

work done = 50 (−0.574)

work done = −28.7 kJ

here we can see work done is negative so its mean work done opposite in direction of inside air

'

7 0
3 years ago
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical
Olin [163]

Answer:

The theoretical density for Niobium is 1.87 g/cm^3.

Explanation:

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density  of the unit cell

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm

1 nm = 10^{-7} cm

On substituting all the given values , we will get the value of 'a'.

\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}

\rho =1.87 g/cm^3

The theoretical density for Niobium is 1.87 g/cm^3.

6 0
3 years ago
Read 2 more answers
Describe the testing process you might use for a new tablet operating system.
rusak2 [61]

Answer:

You may execute some or all Test Cases based on the requirement in all versions of Mobile that is 2g, 3g, and 4g. Tablets and smartphones have special operating-system needs.

Explanation:

3 0
2 years ago
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