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3 years ago

Ughhh my cramps hurt sm

Engineering

2 answers:

kozerog [31]3 years ago

5 0

Answer:

Explanation:Come

tomate

vazorg [7]3 years ago

3 0

Answer:

huh? wdym?

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The driver should be able to see the ground within _____ to the front?

DENIUS [597]

Answer:

12-15 feet

Explanation:

6 0

3 years ago

Realize the function f(a, b, c, d, e) = Σ m(6, 7, 9, 11, 12, 13, 16, 17, 18, 20, 21, 23, 25, 28)using a 16-to-1 MUX with control

NNADVOKAT [17]

Answer:

See explaination

Explanation:

please see attachment for the detailed diagram used in solving the given problem.

It is attached as an attachment.

7 0

3 years ago

A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maxi

saw5 [17]

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, $\tau_{max}$ , is given by the following formula;

$\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )$

$t_w$ = 1 cm = 0.01

h = 29 cm = 0.29 m

$h_w$ = 25 cm = 0.25 m

b = 15 cm = 0.15 m

$I_c$ = The centroidal moment of inertia

$I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )$

$I_c$ = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

$I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}$

$I_c$ = 1.2257083 $\bar 3$ × 10⁻⁴ m⁴

From which we have;

$4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )$

Which gives;

W = 11,416.6879 N

$\sigma _{b.max} = \dfrac{M_c}{I_c}$

$\sigma _{b.max}$ = 1500 N/cm² = 15,000,000 N/m²

$M_c$ = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

$M_{max} = \dfrac{W \cdot L}{4}$

$L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417$

L ≈ 0.64417 m ≈ 64.417 cm.

4 0

3 years ago

An insulated tank having a total volume of 0.6 m3 is divided into two compartments. Initially one compartment contains 0.4 m3 of

IRISSAK [1]

Answer:

a. The final temperature is 69.8°C

b. The final pressure is 2.67bar

c. The amount of entropy produced is 0.38568kJ/K

Explanation:

Pls, check the attached files for detail step by step explanation as typing the solution here can not be explicit.

5 0

4 years ago

Argon is compressed in a polytropic process with n = 1.2 from 100 kPa and 30°C to 1200 kPa in a piston–cylinder device. Determin

gulaghasi [49]

Answer:

181 °C

Explanation:

Initial pressure $P_{1}$ = 100 kPa

Initial temperature $T_{1}$ = 30 °C = 30 + 273 K = 303 K

Final pressure $P_{2}$ = 1200 kPa

Final temperature $T_{2}$ = ?

n = 1.2

For a polytropic process, we use the relationship

( $T_{2}$ / $T_{1}$ ) = ( $P_{2}$ / $P_{1}$ )^γ

where γ = (n-1)/n

γ = (1.2-1)/1.2 = 0.1667

substituting into the equation, we have

( $T_{2}$ /303) = (1200/100)^0.1667

$T_{2}$ /303 = 12^0.1667

$T_{2}$ /303 = 1.513

$T_{2}$ = 300 x 1.513 = 453.9 K

==> 453.9 - 273 = 180.9 ≅ 181 °C

5 0

3 years ago