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liq [111]
4 years ago
12

The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of 0.75 in2, Determine the mag

nitude of the force P needed to displace point B 0.10 in. downward.
Engineering
1 answer:
algol134 years ago
8 0

Answer:

0.029N

Explanation:

Pressure = displacement × density of steel × acceleration due to gravity

displacement = 0.1in = 0.1×0.0254m = 0.00254m, density of steel = 7900kg/m^3, acceleration due to gravity = 9.8m/s^2

Pressure = 0.00254×7900×9.8 = 196.65N/m^2

Force(P) = pressure × area

Area of 1 pin-connected A-36 steel member = 0.75in^2 = 4.84×10^-6m^2

Area of 3 pin-connected A-36 steel member = 3×4.84×10^-6 = 1.452×10^-5m^2

Force(P) = 196.65 × 1.4652×10^-5 = 0.0029N

You might be interested in
Realiza las siguientes conversiones.
amid [387]

Answer:

a) 4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) 21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) 0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) 0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.  

e) 23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

Explanation:

a) <em>4 hectómetros cuadrados a decámetros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un hectómetro cuadrado equivale a 100 decámetros cuadradps. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

x = 4\,Hm^{2}\times\frac{100\,Dm^{2}}{1\,Hm^{2}}

x = 400\,Dm^{2}

4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) <em>21345 centímetros cuadrados a metros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

x = 21345\,cm^{2}\times \frac{1\,m^{2}}{10000\,cm^{2}}

x = 2,135\,m^{2}

21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) <em>0,592 kilómetros cuadrados a metros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 1000000 metros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

x = 0,592\,km^{2}\times \frac{1000000\,m^{2}}{1\,km^{2}}

x = 592000\,m^{2}

0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) <em>0,102 metros cuadrados a centímetros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

x = 0,102\,m^{2}\times \frac{10000\,cm^{2}}{1\,m^{2}}

x = 1020\,cm^{2}

0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.

e) <em>23911 kilómetros cuadrados a hectómetros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 100 hectómetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

x = 23911\,km^{2}\times \frac{100\,Hm^{2}}{1\,km^{2}}

x = 2391100\,Hm^{2}

23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

7 0
4 years ago
A car is moving at 68 miles per hour. The kinetic energy of that car is 5 × 10 5 J.How much energy does the same car have when i
Blababa [14]

Answer:

The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

Explanation:

Given the data in the question;

Initial velocity v₁ = 68 miles per hour = 30.398 meter per seconds

let mass of the car be m

kinetic energy of that car is 5 × 10⁵ J

so

E₁ = \frac{1}{2}mv²

we substitute

5 × 10⁵  = \frac{1}{2} × m × ( 30.398 )²

5 × 10⁵  = \frac{1}{2} × m × ( 30.398 )²

5 × 10⁵ = m × 462.019

m =  5 × 10⁵ / 462.019

m = 1082.2065 kg

Now, Also given that; v₂ = 97 miles per hour = 43.362 meter per seconds

E₂ = \frac{1}{2}mv₂²

we substitute

E₂ = \frac{1}{2} × 1082.2065 × ( 43.362 )²

E₂ = \frac{1}{2} × 1082.2065 × 1880.263

E₂ = 1.017 × 10⁵ J

Therefore, The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

6 0
3 years ago
A boiler is designed to work at 14bar and evaporate 8 kg/s of water. The inlet water to the boiler has a temperature of 400C and
bearhunter [10]

Answer:

Explanation: 2 is thy answer

7 0
3 years ago
A sports car has a drag coefficient of 0.29 and a frontal area of 20 ft2, and is travelling at a speed of 120 mi/hour. How much
Andrej [43]

Answer:

Power required to overcome aerodynamic drag is 50.971 KW

Explanation:

For explanation see the picture attached

4 0
3 years ago
Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh
Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

R_a=e^{-\left ( \frac{T-r}{n}\right )B}

R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}

R_a=e^{-4.4}=0.01227

For B

R_b=e^{-\left ( \frac{300-336}{100}\right )3}

R_b=e^{1.08}=2.944

B is better for 100 hours

(b)For 750 hours

R_a=e^{-0.5866}=0.55621

R_b=e^{0.144}=1.154

So here B is more Reliable.

3 0
3 years ago
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