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3 years ago

7

Type the correct answer in the box. Use numerals instead of words. What’s the mechanical advantage of the lever? Tamara uses a l

ever to pry open a stuck window. She applies 50 newtons of force to one end of the lever. The other end of the lever applies 150 newtons of force to the window. The mechanical advantage of Tamara’s lever is Blank

1 answer:

IrinaK [193]3 years ago

6 0

Answer:3 on PLATO

Explanation:

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On a piece of paper, sketch the x-y stress state and the properly oriented principal stress state. Use the resulting sketch to a

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Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step and very detailed solution of the given problem

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4 years ago

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$

$C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}$

$C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}$

$C_{10} = 30962.449 \ lbf$

Recall that:

1 kN = 225 lbf

∴

$C_{10} = \dfrac{30962.449}{225}$

$\mathbf{C_{10} = 137.611 \ kN}$

7 0

3 years ago

A building wall has dimensions of 3 m tall and 10 m wide. It is constructed of 2 cm. wallboard (k = 0.5 W/m-C) on the inside, 3

Art [367]

Answer: heat loss through wall is 16.58034kW

Temperature of inside wall surface is 47°c

Temperature of outside wall surface is -2.7°c

Explanation:detailed calculation and explanation is shown in the image below.

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3 years ago

A company wants to develop audio speakers using an inexpensive type of plastic that has a very high quality of sound output. Whi

topjm [15]

Answer:A

Explanation:

Those who want to save money and will use the product for only a few years

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4 years ago

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

KIM [24]

Answer:

$\eta$ =0.60

Explanation:

Given :Take $\gamma$ =1.4 for air

$P_1=100 KPa ,T_1=300K$

$\frac{V_1}{V_2}$ =r ⇒ r=16

As we know that

$T_2=T_1(r^{\gamma-1})$

So $T_2=300\times 16^{\gamma-1}$

$T_2$ =909.42K

Now find the cut off ration $\rho$

$\rho=\frac{V_3}{V_2}$

$\frac{V_3}{V_2}=\frac{T_3}{T_2}$

$\rho=\frac{2031}{909.42}$

$\rho=2.23$

So efficiency of diesel engine

$\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}$

Now by putting the all values

$\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}$

So $\eta$ =0.60

So the efficiency of diesel engine=0.60

7 0

3 years ago