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Yakvenalex [24]
3 years ago
7

A screw is a simple machine that can be described as an inclined plane wrapped around a cylinder.

Engineering
1 answer:
Phoenix [80]3 years ago
7 0

Answer:

thanks thanks thanks thanks

Explanation:

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A pressure transducer has the following specifications: Input rage: 0-100 psi and the corresponding Output Range: 0-5 Volts. Lin
motikmotik

Answer:

 0.287

Explanation:

Design-stage uncertainty can be expressed as :

Ud = √ Uo^2 + Uc^2 ------ ( 1 )

where : Uo = 1/2( resolution value ) = 1/2 * 0.01 V = 0.005 V

Uc = √(0.10)^2 + (0.10)^2 + (0.15)^2 + (0.20)^2  = 0.287

back to equation 1

Ud = √ ( 0.005)^2 + ( 0.287 )^2 =  0.287

6 0
3 years ago
Any programmer who writes a Diophantine equation solver must occasionally encounter an infinite loop.
Sergio [31]

Answer: True

Explanation:

An Infinite loop occurs in a computer programming when a sequence of instructions run endlessly without stopping until there is an external intervention, this intervention could be pull or plug. It is also known as endless loop. It occurs due to using of variables that are not properly updated or when there is an error in looping condition.

4 0
4 years ago
Match each context to the type of the law that is most suitable for it.
Bas_tet [7]

Answer:

sorry i dont understand the answer

Explanation:

but i think its a xd jk psml lol

5 0
3 years ago
Carl why is there a dead man in the living room?
scoray [572]
You always need some company
3 0
3 years ago
A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p
Stella [2.4K]

Answer:

Cut-off ratio\dfrac{V_3}{V_2}=6.15

Cxpansion ratio\dfrac{V_4}{V_3}=3.25

The exhaust temperatureT_4=1997.5R

Explanation:

Compression ratio CR(r)=20

\dfrac{V_1}{V_2}=20

P_2=P_3=920 psia

T_1=520 R ,T_{max}=T_3,T_3=3200 R

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{520}=20^{1.4 -1}

T_2=1723.28R

\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}

\dfrac{V_3}{V_2}=\dfrac{3200}{520}

So cut-off ratio\dfrac{V_3}{V_2}=6.15

\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}

Now putting the values in above equation

\dfrac20=\dfrac{V_4}{V_3}\times 6.15

\dfrac{V_4}{V_3}=3.25

So expansion ratio\dfrac{V_4}{V_3}=3.25.

\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}

\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}

T_4=1997.5R

So the exhaust temperatureT_4=1997.5R

3 0
4 years ago
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