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3 years ago

6

Which type of software installation has you install all the most commonly used files and programs to your computer's hard drive?

O full O custom Q common O complete Day

1 answer:

Archy [21]3 years ago

5 0

Custom software installation will be the answer

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The uniform beam is supported by two rods AB and CD that have crosssectional areas of 10 mm2 and 15 mm2 , respectively. Determin

ddd [48]

Answer:

w=2.25

Explanation:

It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.

The surface of the cross-section of the stapes was determined:

A_ab= 10 mm^2

A-cd= 15 mm^2

The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.

σ_ab = F_ab/A_ab $\leq$ σ_allow

σ_cd = F_cd/A_cd $\leq$ σ_allow

In the next step we will determine the static size: Picture b).

We apply the conditions of equilibrium:

∑F_x=0

∑F_y=0

∑M=0

∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0

==> F_cd = 2*w*k*N

∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0

==> F_ab = w*k*N

Now we determine the load w

<u>Sector AB: </u>

σ_ab = F_ab/A_ab $\leq$ σ_allow=300 KPa

= w/10*10^-6 $\leq$ σ_allow=300 KPa

w_ab = 3*10^-3 kN/m

<u>Sector CD: </u>

σ_cd = F_cd/A_cd $\leq$ σ_allow=300 KPa

= 2*w/15*10^-6 $\leq$ σ_allow=300 KPa

w_cd = 2.25*10^-3 kN/m

w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}

==> w=2.25 * 10^-3 kN/m

<u>The solution is: </u>

w=2.25 N/m

note:

find the attached graph

6 0

3 years ago

What 2 disciplines comprise systems engineering?

kirill [66]

Answer:

Engineering and the Engineering management

Explanation:

Systems engineering is a discovery process that includes managing and designing complex systems to formalize the various new research approach or methods.

It is an interdisciplinary holistic approach or field that is comprised of <u>engineering and the engineering management.</u>

The three types of systems engineering are: Product Systems Engineering, Enterprise Systems Engineering and Service Systems Engineering.

5 0

3 years ago

A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

6 0

4 years ago

Write a function "funthree" that will print a box of characters. The function will always receive as the first input argument th

stiv31 [10]

Answer: i8g7ieusgr7eytuu7esieso87 sugr7pi8y8hiuh9yehroe998 rydyh9 t9ry9 7 fdgerje 78re87es 7yehr87ehtu9hu7b puuhihugogi;ghi uhugyug fhglhfu fufbiup hughghuihu fhuihihiuhg uhfuhuig8fguh hguihfhjliigihfuhf;gjihgh hfuyt8uyiiohiodohi

Explanation: so first you are going to hyigfgegidii9thdf5dh8yy7 gdiuigp 87hgiuf 8yhijh gihoighhgd

4 0

4 years ago

On some late model cars, climate control systems wil create trouble codes for trouble shooting purposes.

Advocard [28]

I think it’s true because I mean that where you have air conditioner and stuff but as though it’s probably not codes but in a car there probably will be codes

8 0

3 years ago