Why has the study of genetic differences in intelligence been so controversial?

A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal

melamori03 [73]

Answer:

Therefore, the force supported by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

$\sum M__A }=0$

- 40 (180) - (80) $F_E$ Cos 37° + 55

-7200 + $F_E$ (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79 $F_E$

- 30.79 $F_E$ = 7200

$F_E$ = $-\frac{7200}{30.79}$

$F_E$ = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

$\sum f_y = 0$

$F_E$ sin 37 + $F_A$ cos 20 - $F_G$ = 0

- 233.84 sin 37 + $F_A$ cos 20 - 180 = 0

$F_A$ cos 20 = 320.73

$F_A$ = $\frac{320.73}{cos 20}$

$F_A$ = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

$\sum f_y = 0$

$F_E$ cos 37 - $F_{CB} + F_A$ sin 20° = 0

-233.84 cos 37 - $F_{CB}$ + 341.31 sin 20° = 0

-186.75 - $F_{CB}$ + 116.74 = 0

- $F_{CB}$ -70.01 = 0

- $F_{CB}$ = 70.01

$F_{CB}$ = - 70.01 lb

Therefore, the force supported by the hydraulic cylinder is = -70.01 lb

7 0

4 years ago

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

6 0

3 years ago

At what temperature is Fahrenheit scale reading equal to twice of celsius?

gayaneshka [121]

At -24.613 F<span> temperature is the Fahrenheit scale reading equal to twice that of the Celsius and at </span>160 F<span> temperatuar is the Fahrenheit scale reading equal to half that of the Celsius? So, 160 degrees Celsius will equal 320 degrees Fahrenheit.</span>

5 0

3 years ago

Write a use of magnetic force and frictional force each

Firdavs [7]

Magnetic force:Computer hard drives use magnetism to store the data on a rotating disk. More complex applications include: televisions, radios, microwave ovens, telephone systems, and computers. ... Frictional force:Quite often uses of friction can be seen from how things would be without friction.

7 0

3 years ago

A/an ___ is a machine which tells us about the strength and speed of sisemec waves.

svp [43]

The answer is Seismograph

3 0

3 years ago

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