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xeze [42]
3 years ago
12

- if `check_1` and `check_2` variables are both True, it should set the value of a variable `outcome` to the string 'BOTH' - eli

f `check_1` is True and `check_2` is False, it should set the value of a variable `outcome` to the string 'ONE' - elif `check_1` is False and `check_2` is True, it should set the value of a variable `outcome` to the string 'TWO' - else (meaning both must be False), it should set the value of a variable `outcome` to the string 'NEITHER'
Engineering
1 answer:
tresset_1 [31]3 years ago
4 0

Answer:

See Explaination

Explanation:

if(check1 and check2):

outcome = "BOTH"

elif(check1):

outcome = "ONE"

elif(check2):

outcome = "TWO"

else:

outcome = "NEITHER"

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In surveying , supposing we can not pull the tape because it is passing through a shallow river. What will i do to obtain an acc
natulia [17]

Answer:

Hold the tape in place and go down to the end.

Explanation:

7 0
2 years ago
A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta
Vikki [24]

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore\theta angle = 0

and radial component of given velocity is zero

we haveh = r_o v_r_o = 6378+600 =6.97*10^6 m

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)

GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s

so

\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)

solvingt for \epsilon)

\epsilon = 0.22)

therefore eccentrcity of orbit is 0.22

6 0
3 years ago
10–25. The 45° strain rosette is mounted on the surface of a shell. The following readings are obtained for each gage: ε a = −20
vazorg [7]

Answer:

The answer is 380.32×10^-6

Refer below for the explanation.

Explanation:

Refer to the picture for brief explanation.

7 0
3 years ago
Select a research proposal topic that relates to electrical and electronics engineering and write a proposal report taking into
fiasKO [112]

Expand your technical knowledge, form global networks and balance life & work commitments. Our advanced diplomas remain current with technological and industry developments.

8 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

8 0
4 years ago
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