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3 years ago

7

What is 1 lb to 4.45 N (newtons)?

1 answer:

algol133 years ago

6 0

Answer:

4.44822

Explanation:

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The "lead" in pencils is a graphite composition with a Young's modulus of about 1 ✕ 10⁹ N/m². Calculate the change in length (in

sweet [91]

Answer:

0.701 mm.

Explanation:

From Hook's law,

Young modulus = Stress/Strain.

Stress = Force/area = F/A

Strain = ΔL/L.

Therefore,

γ = (F/A)/(ΔL/L)

γ = FL/ΔLA ...................... Equation 1

Where γ = Young's modulus, F = Force, L = Length of the pencil, ΔL = change in length of the pencil, A = cross sectional area.

Make ΔL the subject of the equation,

ΔL = FL/γA......................... Equation 2

But,

A = πd²/4......................... Equation 3

Where d = diameter

Substitute equation 3 into equation 2

ΔL = 4FL/γπd²................ Equation 4

Given: F = 4.9 N, L = 55 mm = 0.055 m, d = 0.7 mm = 0.0007 m, γ = 1×10⁹ N/m², π = 3.14.

Substitute into equation 4

ΔL = (4×4.9×0.055)/(1×10⁹×3.14×0.0007²)

ΔL = 1.078/1538.6

ΔL = 7.01×10⁻⁴ m.

ΔL = 0.701 mm.

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3 years ago

How do you measure the volume of irregular object? explain with diagram

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Answer:

by using graphical meythod

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3 years ago

Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th

Yuki888 [10]

Complete Question

Diagram for this shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

The density `of wooden handle is $\rho_w = 860 kg/m^3$

The density `of soft-metal head is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

$m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is mathematically obtain with the formula

$A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter of the wooden handle which from the diagram is

$27mm = \frac{27}{1000} = 0.027m$

So $A_w = \frac{\pi}{4} * 0.027^2$

$l_w$ is the length of the the wooden handle which is given in the diagram as $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

$m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

$= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

$m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is mathematically obtain with the formula

$A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter of the soft-metal head which from the diagram is

$36mm = \frac{36}{1000} = 0.036m$

So $A_s = \frac{\pi}{4} * 0.036^2$

$l_s$ is the length of the the soft-metal head which is given in the diagram

as $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass

$m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

$=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

$(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$

Substituting values

$(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

$=5.12*10^{-3}kg \cdot m^2$

Generally the mass moment of inertia about x-axis for the soft-metal head is

$(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

$b=l_w +\frac{d_s}{2}$

Substituting values

$b = 0.315 + \frac{0.036}{2}$

$= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

$(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

$=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

$I_{xx} = (I_{xx})_w + (I_{xx})_s$

$= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

$I_{xx} = 5.12*10^{-3} +0.113$

$I{xx}= 0.119 kg \cdot m^2$

8 0

3 years ago