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2 years ago

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 12.0K .

Physics

1 answer:

Bingel [31]2 years ago

3 0

Answer: a) - 437.8° F, b) - 261°c.

Explanation: a) the kelvin and Fahrenheit temperature scale are related by the formulae below.

5 (°F - 32) = 9 (k - 273)

Where °F = temperature in Fahrenheit and k = temperature in kelvin.

For question A, k = 12.0, by substituting to have the value for °F, we have

5(°F - 32) = 9 ( 12 - 273)

5(°F - 32) = 9(-261)

5(°F - 32) = - 2349

°F - 32 = - 2349/5

°F - 32 = - 469.8

°F = - 469.8 + 32

°F = - 437.8

Question B

The centigrade and kelvin scale are related by the formulae below

°c = k - 273

Where °c = temperature in centigrade and k = temperature in kelvin =12

°c = 12 - 273

°c = - 261

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Comparativo entre a máquina a vapor de Neu carmen e James Watt!

Bas_tet [7]

Answer:

jame watt

Explanation:

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3 years ago

Water waves in a small tank are 6.6 m long. They travel at a speed of 2.9 m/s. What is the frequency of the water

liubo4ka [24]

Answer:

.439 Hz

Explanation:

Wavelength = 6.6 m

speed = 2.9 m/s

2.9 = 6.6 f

f = 2.9/6.6 = .439 Hz

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1 year ago

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

3 years ago

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

$\sigma_h = \frac{Pd}{2t}$

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

$\sigma_l = \frac{Pd}{4t}$

The letters have the same meaning as before.

Then he hoop stress would be,

$\sigma_h = \frac{Pd}{2t}$

$\sigma_h = \frac{75 \times 8}{2\times 0.25}$

$\sigma_h = 1200psi$

And the longitudinal stress would be

$\sigma_l = \frac{Pd}{4t}$

$\sigma_l = \frac{75\times 8}{4\times 0.25}$

$\sigma_l = 600Psi$

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

$\tau_{max} = \frac{\sigma_h}{2}$

$\tau_{max} = \frac{1200}{2}$

$\tau_{max} = 600Psi$

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0

2 years ago

Listed below are four activities. In which activity is friction useful?

Likurg_2 [28]

Answer:

a. climbing up a cliff

Explanation:

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2 years ago