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3 years ago

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A sheet of aluminum (al) foil has a total area of 1.000 ft2 and a mass of 3.716 g. what is the thickness of the foil in millimet

ers (density of al = 2.699 g/cm3)?

1 answer:

Andrews [41]3 years ago

3 0

The mass of the sheet of aluminium is the product between its density d and its volume V:
$m=dV$
Re-arranging the equation, we can find the volume of the sheet:
$V= \frac{m}{d}= \frac{3.716 g}{2.699 g/cm^3}= 1.377 cm^3$

Converted in millimeters,
$V=1377 mm^3$

The area is
$A=1.00 ft^2$
Since $1 ft = 0.305 m$ , this area corresponds to
$A=1.00 ft^2 = (0.305 m)^2 = 0.093 m^2 = 0.093 \cdot 10^6 mm^2$

So we can now find the thickness of the sheet of aluminium:
$t= \frac{V}{A}= \frac{1377 mm^2}{0.093 \cdot 10^6 mm^2}=0.015 mm$

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The gravitational force between two objects that are 2.1 times 10^-1 m apart is 3.2 times 10^-6 N. If the mass of one object is

Shtirlitz [24]

Answer:

Mass of the other object is 38.45 kg.

Explanation:

Given:

The gravitational force between two objects is, $F=3.2\times 10^{-6}$ N

Mass of one object is, $m_{1}=55\ kg$

Distance between the objects is, $r=2.1\times 10^{-1}\ m$

Gravitational constant is, $G =6.674\times 10^{-11}\ m^3 kg^{-1}s^{-2}$

Let the mass of the other object be $m_{2}\ kg$

Gravitational force is given as:

$F=\frac{Gm_1m_2}{r^2}$

Plug in the given values and solve for $m_2$ . This gives,

$3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{(2.1\times 10^{-1})^2}\\3.2\times 10^{-6}=\frac{6.674\times 10^{-11}\times 55\times m_2}{0.0441}\\3.2\times 10^{-6}=8.3236\times 10^{-8}m_2\\m_2=\frac{3.2\times 10^{-6}}{8.3236\times 10^{-8}}= 38.45\ kg$

Therefore, the mass of the other object is 38.45 kg.

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4 years ago

What is the dimension of area

never [62]

Thus, area is the product of two lengths and so has dimension L2, or length squared.

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A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing

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Given:

Shaft Power, P = 7.46 kW = 7460 W

Speed, N = 1200 rpm

Shearing stress of shaft, $\tau _{shaft}$ = 30 MPa

Shearing stress of key, $\tau _{key}$ = 240 MPa

width of key, w = $\frac{d}{4}$

d is shaft diameter

Solution:

Torque, T = $\frac{P}{\omega }$

where,

$\omega = \frac{2\pi N}{60}$

$T = \frac{7460}{\frac{2\pi (1200 )}{60}}$ = 59.365 N-m

Now,

$\tau _{shaft} = \tau _{max} = \frac{2T}{\pi (\frac{d}{2})^{3}}$

$30\times 10^{6} = \frac{2\times 59.365}{\pi (\frac{d}{2})^{3}}$

d = 0.0216 m

Now,

w = $\frac{d}{4}$ = $\frac{0.02116}{4}$ = 5.4 mm

Now, for shear stress in key

$\tau _{key}$ = $\frac{F}{wl}$

we know that

T = $F \times r$ = F. $\frac{d}{2}$

⇒ $\tau _{key}$ = $\frac{\frac{T}{\frac{d}{2}}}{wl}$

⇒ $240\times 10^{6}$ = $\frac{\frac{59.365}{\frac{0.0216}{2}}}{0.054l}$

length of the rectangular key, l = 4.078 mm

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