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soldi70 [24.7K]
3 years ago
7

what are some questions about simple machines that don't make me sound like i didn't pay attention in the lesson?

Engineering
1 answer:
Margaret [11]3 years ago
8 0

Answer:

what are simple machines?

Explanation:

it is 2020 let's be honest all

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A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l
Papessa [141]

Answer:

(a) 2.5 ksi

(b) 0.1075 in

Explanation:

(a)

E=\frac {\sigma}{\epsilon}

Making \sigma the subject then

\sigma=E\epsilon

where \sigma is the stress and \epsilon is the strain

Since strain is given as 0.025% of the length then strain is \frac {0.025}{100}=0.00025

Now substituting E for 10\times 10^{6} psi then

\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi

(b)

Stress, \sigma= \frac {F}{A} making A the subject then

A=\frac {F}{\sigma}

A=\frac {\pi(d_o^{2}-d_i^{2})}{4}

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that 2t=d_o - d_i where t is thickness

Now substituting

\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}

\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4

(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

But the outer diameter is given as 2 in hence

(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}

d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in

As already mentioned, 2t=d_o - d_i hence t=0.5(d_o - d_i)

t=0.5(2-1.785)=0.1075 in

3 0
3 years ago
A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container
kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

delta S_{12}=\int\limits^2_1  \, \frac{dQ}{T}

In an isobaric process heat (Q) is defined as:

Q= m*Cp*dT

Replacing in the equation for entropy  

delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:  

delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}  

Solving the integral we get the expression to estimate the entropy change in the system  

delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is 0.85\frac{kJ}{Kg*K}

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

Q= m*Cp*dT

With dt=T_{2}-T_{1} clearing for T2 we get:

T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K

Now we can estimate the entropy change in the system

delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.  

b) see picture.

3 0
3 years ago
P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame
eimsori [14]

Answer:

a)  m=0.17kg/s

b)  Ma=0.89

Explanation:

From the question we are told that:

Pressure P=60kPa

Diameter d=3cm

Generally at sea level

T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4

Generally the Power series equation for Mach number is mathematically given by

\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}

\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}

Ma=0.89

Therefore

Mass flow rate

\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\rho=0.848kg/m^3

Generally the equation for Velocity at throat is mathematically given by

V=Ma(r*T_0\sqrt{T_e})

Where

T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}

T_e=248

Therefore

V=0.89(1.4*288\sqrt{248})\\\\V=284

Generally the equation for Mass flow rate is mathematically given by

m=\rho*A*V

m=0.84*\frac{\pi}{4}*3*10^{-2}*284

m=0.17kg/s

6 0
2 years ago
Engineered lumber should not be used for
Dimas [21]

Answer:

Composite panel garage doors

Explanation:

8 0
2 years ago
The correct statement about the lift and drag on an object is:_______
Lisa [10]

Answer:

(a). the resultant force in the direction of the freestream velocity is termed the drag and the resultant force normal to the freestream velocity is termed the lift

Explanation:

When a fluid flows around the surface of an object, it exerts a force on it. This force has two components, namely lift and drag.

The component of this force that is perpendicular (normal) to the freestream velocity is known as lift, while the component of this force that is parallel or in the direction of the fluid freestream flow is known as drag.

Lift is as a result of pressure differences, while drag results from forces due to pressure distributions over the object surface, and forces due to skin friction or viscous force.

Thus, drag results from the combination of pressure and viscous forces while lift results only from the<em> pressure differences</em> (not pressure forces as was used in option D).

The only correct option left is "A"

(a). the resultant force in the direction of the freestream velocity is termed the drag and the resultant force normal to the freestream velocity is termed the lift

8 0
3 years ago
Read 2 more answers
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