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2 years ago

According to the

Engineering

1 answer:

zysi [14]2 years ago

3 0

Answer:

The part of the system that is considered the resistance force is;

Explanation:

The simple machine is a system of pulley that has two pulleys

The effort, which is the input force at A gives the value of the tension at C and D which are used to lift the load B

Therefore, we have;

A = C = D

B = C + D = C + C = 2·C

∴ C = B/2

We have;

C = B/2 = A

Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B

The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done

It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force

Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.

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To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr

alukav5142 [94]

$\ T_{c}$ has greater effect.

Explanation:

$\eta_{\max }=1-\frac{T_{c}}{T_{A}}$

$T_{c}\\$ = Temperature of cold reservoir

$T_{H}$ = Temperature of hot reservoir

when $T_{c}$ is decreased by 't',

$\eta_{\text {incre }}$ = $1-\frac{\left(\tau_{c}-t\right)}{T_{H}}$

$=n \ + \frac{t}{T_{n}}$ $-(i)$

when ${T_{H}}$ is increased by 'T'

$\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)$

$\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}$

7 0

2 years ago

weight of 1000 pounds is suspended from two cables. The allowable stress in the cables is 1500 psi. Find the minimum diameter fo

kari74 [83]

Answer:

The minimum diameter for each cable should be 0.65 inches.

Explanation:

Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,

(1/2)(Weight/Cross Sectional Area) = Allowable Stress

Here,

Weight = 1000 lb

Cross-sectional area = πr²

where, r = minimum radius for each cable

(1/2)(1000 lb/πr²) = 1500 psi

500 lb/1500π psi = r²

r = √1.061 in²

r = 0.325 in

Now, for diameter:

Diameter = 2(radius) = 2r

Diameter = 2(0.325 in)

Diameter = 0.65 in

7 0

2 years ago

Which of the following best distinguishes between superficial design improvements and functional

Contact [7]

Answer:

Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.

Explanation:

As a PC board designer, I would sometimes spend a certain amount of time making traces have shorter routes, or fewer layer changes or bends. (I wanted to make the layout "pretty.") In some cases, these changes are superficial, affecting the appearance only. In some cases, they are functional, reducing crosstalk or emissions or susceptibility to interference.

I deal with a web site that seems to be changing all the time (Brainly). In many cases, the same information is rearranged on the page—a superficial change. In other cases, the information being displayed changes, or the way that certain information is accessed changes. These are functional changes. (Sometimes, they "enhance performance," and sometimes they don't, IMO.)

In short ...

Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.

8 0

2 years ago

An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$