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2 years ago

A heating torch is usually referred to as what? Stick Flower Rose-bud Lighter

Engineering

2 answers:

Aleonysh [2.5K]2 years ago

3 0

I would select rose-bud.

makvit [3.9K]2 years ago

3 0

Answer:

Rose-bud

Explanation:

Here is a picture of one :)

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Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

3 years ago

An operating gear box (transmission) has 350 hp at its input shaft while 250. hp are delivered to the output shaft. The gear box

True [87]

Answer:

Rate of Entropy =210.14 J/K-s

Explanation:

given data:

power delivered to input = 350 hp

power delivered to output = 250 hp

temperature of surface = 180°F

rate of entropy is given as

$Rate\ of\ entropy = \frac{Rate\ of \ heat\ released}{Temperature}$

T = 180°F = 82°C = 355 K

Rate of heat = (350 - 250) hp = 100 hp = 74600 W

Rate of Entropy $= \frac{74600}{355} = 210.14 J/K-s$

8 0

3 years ago

We are given a CSP with only binary constraints. Assume we run backtracking search with arc consistency as follows. Initially, w

sweet-ann [11.9K]

We are given a CSP with only binary can concentrate assume we run backtrackingSearch with ARC

8 0

2 years ago

Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br

vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation $v = v_o + a_ct$ to determine the velocity of car B.

where;

$v_o =$ initial velocity

$a_c$ = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

$S = d + v_ot + \dfrac{1}{2}at^2$

Then:

$v_B = 60-12t$

The distance traveled by car B in the given time (t) is expressed as:

$S_B = d + 60 t - \dfrac{1}{2}(12t^2)$

For car A, the needed time (t) to come to rest is:

$v_A = 60 - 18(t-0.75)$

Also, the distance traveled by car A in the given time (t) is expressed as:

$S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

Relating both velocities:

$v_B = v_A$

$60-12t = 60 - 18(t-0.75)$

$60-12t =73.5 - 18t$

$60- 73.5 = - 18t+ 12t$

$-13.5 =-6t$

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

$S_B = S_A$

$d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

$d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2$

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

3 0

3 years ago

5 cause of a electrical problem

cupoosta [38]

Answer:

1. Poor circuit protection

2.Grounding issue

3. lighting problem

4. Electrical shocks

5. High electricity bills

Explanation:

8 0

3 years ago