What is the command in the java script that receives data from the user

A series of experiments is conducted in which a thin plate is subjected to biaxial tension/compression, σ1, σ2 , the plane surfa

Andrew [12]

Answer:

Explanation:

3 0

3 years ago

A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at

DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

$dQ = m \times c_p \times (T_2 -T_1)$

For ideal gas, $c_p$ = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

3 0

3 years ago

What oil specification is used to express the thickness (viscosity) of the oil?

Marianna [84]

Answer:

Viscosity is notated using the common classification “XW-XX”. The number preceding the “W” (winter) rates the oil's flow (viscosity) at zero degrees Fahrenheit (-17.8 degrees Celsius). The lower the number, the less the oil thickens in cold weather.

7 0

2 years ago

The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r

Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

$K = \sigma Y \sqrt{\pi a}$

where;

fracture toughness K = 137 MPa $m^{1/2}$

geometry factor Y = 1

applied stress $\sigma$ = ???

crack length a = 2mm = 0.002

∴

$137 =\sigma \times 1 \sqrt{ \pi \times 0.002 }$

$137 =\sigma \times 0.07926$

$\dfrac{137}{0.07926} =\sigma$

$\sigma = 1728.489 MPa$

Now, the tensile impact obtained is:

$\sigma = \dfrac{P}{A}$

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0

3 years ago

The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both

DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

L^2 = x_a^2 + y_b^2 ....... 1

y_b = sqrt( 20^2 - 16^2 )

y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

0 = 2*x_a*v_a + 2*y_b*v_b

0 = x_a*v_a + y_b*v_b ..... 2

0 = 16*36 + 12*v_b

v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

0 = 9 + 4*a_a/3 + 16 + a_b

4*a_a/3 + a_b = -25

4*a_a + 3*a_b = -75 .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

F_x = m_a*a_a

P - R_r*16/20 = m_a*a_a

- For Slider B, Apply Newton's second law of motion in y direction:

F_y = m_b*a_b

- R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

P - 4*m_b*a_b / 3 = m_a*a_a

3P = 3*m_a*a_a + 4*m_b*a_b ... 4

- Where, P = Is the force acting on slider A

P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.

5 0

3 years ago