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tatuchka [14]
3 years ago
7

Calculate the maximum load that a 7075 series aluminum alloy bar (with a T6 temper heat treatment) can support without permanent

ly deforming. The bar has a square cross section whose edge length is 8.1 mm.
Engineering
1 answer:
Aleksandr [31]3 years ago
6 0

Answer:

The maximum load the bar can withstand = 35.43 KN

Explanation:

Ultimate tensile strength of the given aluminium bar \sigma = 540 M pa

Cross section area of the bar = 8.1^{2}  = 65.61 mm^{2}

We know that the ultimate strength of the bar is calculated from

\sigma = \frac{P_{max} }{A}

540 = \frac{P_{max} }{65.61}

P_{max} = 540 × 65.61

P_{max} =  35.43 KN

Therefore the maximum load the bar can withstand = 35.43 KN

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ahrayia [7]

Answer:

a clock

Explanation:

you use a clock in water

3 0
2 years ago
A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I
Liula [17]

Answer:

  • fire brick / common brick : 1218 °C
  • common brick / magnesia : 1019 °C
  • magnesia / steel : 90.06 °C
  • heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

  R = d/k

so the thermal resistances of the layers of furnace wall are ...

  R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

  R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

  R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

  R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

  R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

__

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

__

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

  fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

  1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

  common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

  magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0
3 years ago
In C++ the declaration of floating point variables starts with the type name float or double, followed by the name of the variab
zubka84 [21]

Answer:

The given grammar is :

S = T V ;

V = C X

X = , V | ε

T = float | double

C = z | w

1.

Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.

From the given grammar,

Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.

No other variables generate variable X or ε.

So, only variable X is nullable.

2.

First of nullable variable X is First (X ) = , and ε (epsilon).

L.H.S.

The first of other varibles are :

First (S) = {float, double }

First (T) = {float, double }

First (V) = {z, w}

First (C) = {z, w}

R.H.S.

First (T V ; ) = {float, double }

First ( C X ) = {z, w}

First (, V) = ,

First ( ε ) = ε

First (float) = float

First (double) = double

First (z) = z

First (w) = w

3.

Follow of nullable variable X is Follow (V).

Follow (S) = $

Follow (T) = {z, w}

Follow (V) = ;

Follow (X) = Follow (V) = ;

Follow (C) = , and ;

Explanation:

4 0
3 years ago
Hareem and Shahad are on a road trip to Florida. They pull over to get gas, and as you may know it is sold by the gallon in the
Alona [7]

Answer:

53.3

Explanation:

4×14=56

56÷1.05=53.3

4 0
3 years ago
A battery is connected to a resistor. Increasing the resistance of the resistor will __________. A battery is connected to a res
belka [17]

Answer: the increase in the external resistor will affect and decrease the current in the circuit.

Explanation: A battery has it own internal resistance, r, and given an external resistor of resistance, R, the equation of typical of Ohm's law giving the flow of current is

E = IR + Ir = I(R + r)........(1)

Where IR is the potential difference flowing in the external circuit and Or is the lost voltage due to internal resistance of battery. From (1)

I = E/(R + r)

As R increases, and E, r remain constant, the value (R + r) increases, hence the value of current, I, in the external circuit decreases.

8 0
3 years ago
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