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2 years ago

10

How to determine the center of gravity of a lamina

1 answer:

goldenfox [79]2 years ago

7 0

Take a lamina with three holes near the periphery of the lamina, now suspend the lamina through them, one by one. Draw a line of equilibrium for each suspension point. The point of intersection of these three lines would be the centre of gravity.

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You see an image of yourself in a mirror because light waves are A) diffracted. B) reflected back at you. C) refracted through a

kondaur [170]

Answer:

(b) reflected back at you

Explanation:

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4 years ago

Read 2 more answers

What would changing the frequency of a wave do to the wave?

nata0808 [166]

The data convincingly show that wave frequency does not affect wave speed. An increase in wave frequency caused a decrease in wavelength while the wave speed remained constant. The last three trials involved the same procedure with a different rope tension.

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3 years ago

At the beach in San Francisco (0 meters) the pressure of the atmosphere is 101.325 kPa

Korvikt [17]

Answer:

$P = -\frac{17978}{1609344}(h)+101.325$

Explanation:

Given

$h = height$

$P = Pressure$

$(h_1,P_1) = (0,101.325)$

$(h_2,P_2) = (1609.344 ,83.437 )$

Required

Determine the linear equation for P in terms of h

First, we calculate the slope/rate (m);

The following formula is used:

$m = \frac{P_2 - P_1}{h_2 - h_1}$

Substitute values for P's and h's

$m = \frac{83.347 - 101.325}{1609.344- 0}$

$m = \frac{-17.978}{1609.344}$

$m = -\frac{17.978}{1609.344}$

Multiply by 1000/1000

$m = -\frac{17.978 * 1000}{1609.344*1000}$

$m = -\frac{17978}{1609344}$

The equation is then calculated using:

$P - P_1 = m(h - h_1)$

Substitute values for m, h1 and P1

$P - P_1 = m(h - h_1)$

$P - 101.325 = -\frac{17978}{1609344}(h - 0)$

$P - 101.325 = -\frac{17978}{1609344}(h)$

Make P the subject

$P = -\frac{17978}{1609344}(h)+101.325$

<em>The above is the required linear equation</em>

8 0

3 years ago

Calculate the energy (in eV/atom) for vacancy formation in some metal, M, given that the equilibrium number of vacancies at 296o

Schach [20]

Explanation:

The given data is as follows.

Temperature of metal = $296^{o}C$ = (296 + 273) K

= 569 K

Density of the metal = 8.85 $g/cm^{3}$ = $8.85 \times 10^{-6} g/m^{3}$ (as $1 cm^{3} = 10^{-6} m^{3}$ )

Atomic mass = 51.40 g/mol

Vacancies = $9.19 \times 10^{23} m^{-3}$

Formula to calculate the number of atomic sites is as follows.

n = $\frac{\rho \times N_{A}}{\text{atomic weight}}$

= $\frac{8.85 \times 10^{-6} \times 6.022 \times 10^{23}}{51.40 g/mol}$

= $1.036 \times 10^{17} atom/m^{3}$

Now, we will calculate the energy as follows.

E = $-KT \times ln (\frac{\text{no. of vacancies}}{\text{no. of atomic sites}})$

where, K = $8.62 \times 10^{-5}$

E = $-8.62 \times 10^{-5} \times 569 K \times ln (\frac{9.19 \times 10^{23}}{1.036 \times 10^{17} atom/m^{3}})$

= $78.46 eV/atom$

Therefore, we can conclude that energy (in eV/atom) for vacancy formation in given metal, M, is $78.46 eV/atom$ .

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4 years ago

Diego rivera's mural for the lobby of the rca building was destroyed because

forsale [732]

It included a picture of Vladimir Lenin

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3 years ago