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3 years ago

Explain the different methods that can be used to model the motion of an object.

Physics

1 answer:

aleksandr82 [10.1K]3 years ago

3 0

Answer:Even things that appear to be at rest move.

When we describe the motion of one object with

respect to another, we say that the object is moving

relative to the other object.

• A book that is at rest, relative to the table it lies

on, is moving at about 30 kilometers per second

relative to the sun.

• The book moves even faster relative to the center

of our galaxy.

Explanation:

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Determine the shortest frequency of light required to remove an electron from a sample of ti metal if the binding energy of tita

Mama L [17]

The correct answer is: shortest frequency = 7.86*10^15 $s^{-1}$

Explanation:
The binding energy of titanium = 3.14*10^6 J/mol
The energy required to remove an electron = (3.14*10^6) /(6.023*10^23) = 5.213*10^-18 J
Where 6.023*10^23 = Avagadro number

Since E = hv

Frequency = v = E/h
E = Energy = 5.213*10^-18
h = Planck's constant = 6.626*10^-34
v = (5.213*10^-18) / 6.626*10^-34)
v = 7.86*10^15 $s^{-1}$ (shortest frequency)

7 0

4 years ago

A cantilever beam with a width b=100 mm and depth h=150 mm has a length L=2 m and is subjected to a point load P =500 N at B. Ca

DerKrebs [107]

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

$I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4$

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

$\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa$

Calculate the first moment of area below point C

$Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm$

Now calculate shear stress at point C

$=\frac{FQ}{It}$

$=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa$

Calculate the principal stress at point C

$\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa$

Calculate the maximum shear stress at piont C

$\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2} \\\\=336.28KPa$

6 0

3 years ago

7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr

Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

$d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m$