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Elanso [62]
4 years ago
14

In a Brinell hardness test, a 1500-kg load is pressed into a specimen using a 10-mm-diameter hardened steel ball. The resulting

indentation has a diameter = 3.2 mm. (a) Determine the Brinell hardness number for the metal. (b) If the specimen is steel, estimate the tensile strength of the steel.
Physics
1 answer:
Bess [88]4 years ago
7 0

Answer:

a) BHN = 181.606

b) UTS = 635.621 MPa

Explanation:

given,

Load = 1500 kg

steel ball of diameter (D) = 10 mm

resulting indentation diameter (d)= 3.2 mm

brinell hardness number

BHN = \dfrac{2P}{(\pi D)(D - \sqrt{D^2-d^2})}

BHN = \dfrac{2\times 1500}{(\pi \times 10)(10 - \sqrt{10^2-3.2^2})}

BHN = \dfrac{3000}{16.519}

BHN = 181.606

b) The tensile strength of the steel

UTS = 3.5 (BHN)

UTS = 3.5  x (181.606)

UTS = 635.621 MPa

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When you push a 2.00 kg book resting on a tabletop it takes 4.60 N to start the book sliding. What is the coefficient of static
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The coefficient of static friction is 0.234.

Answer:

Explanation:

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The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate
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Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

L = 10log_{10}\frac{I}{I_{o}}          

I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}      

L' = 10log_{10}\frac{I'}{I_{o}}

I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}        

where

I_{o} = 10^{- 12} W/m^{2}

I = Intensity of sound

Now,

I = \frac{P}{4\pi R^{2}}

Similarly,

I' = \frac{P}{4\pi (R + 11.0)^{2}}

Now,

\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}

\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}

R ^{2} + 22R + 121 = 11.64R^{2}}

10.64R ^{2} - 22R - 121 = 0

Solving the above quadratic eqn, we get:

R = 4.56 m

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