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3 years ago

How is the caliber of a rifle or handgun determined

1 answer:

4 0

Caliber usually is measured as the diameter of the bore from land to opposite land and is expressed in hundredths of an inch, thousandths of an inch, or millimeters. For example, a . 270-caliber rifle bore measures 270/1000ths of an inch in diameter between the lands and has a larger bore diameter than a .

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How to tie shoe please

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2 years ago

A Simply supported wood beam with overhang is subjected to uniformly distributed load q. The beam has a rectangular cross sectio

irinina [24]

Answer:

q = 61.71 KN/m

Explanation:

We know that shear force at one end of the beam is;

F = wl/2

Where;

w is the uniformly distributed load and l is the span.

Thus, in this question, q is the distributed load, so;

F = ql/2

Area of beam section = breadth x depth

In this case,

Area = 200 × 250 = 50000 mm²

We are given allowable shear stress of τa=1.8MPa. This can also be written as τa = 1.8 N/mm²

We know that formula for average shear stress is;

τ_avg = Force/Area

Thus, Force = τ_avg x Area

However, we are given maximum allowable shear stress as 1.8and we know that; τ_max = 1.5 × τ_avg

Thus, τ_avg = 1.8/1.5 = 1.2

Hence;

Force = 1.2 × 50,000 = 60000 N

We need

So from the earlier equation F = ql/2,we can get; 60000 = ql/2

ql = 120000 - - - - - (1)

Now, to the bending stress, we know that section modulus of a rectangular section is;

Z = bd²/6

So,for this question, we have;

Z = (200 × 250²)/6

Z = 2083333.33 mm²

Maximum bending moment of a simply supported beam is wl²/8

So,in this case, M = ql²/8

So,formula for maximum bending stress = M/Z

So, plugging in the values, we have ;

σ_max = (ql²/8) / 2083333.33

We are given σ= 14 MPa or 14 N/mm²

Thus;

14 = (ql²/8) / 2083333.33

ql² = 14 × 2083333.33 × 8

ql² = 233333332.96 - - - eq(2)

From equation 1,we saw that;ql = 120000.

Putting this for ql in equation 2,we will get;

120000l = 233333332.96

l = 233333332.96/120000

l = 1944.44 mm

So from eq 1,q = 120000/l

q = 120000/1944.44

q = 61.71 KN/m

6 0

3 years ago

In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at

Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

$V = 690V$