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2 years ago

8

What happens to particles of a substance as its temperature increases? A the average kinetic energy increases B the average kine

tic energy decreases C the average kinetic energy stays the same D nothing happens

1 answer:

Gemiola [76]2 years ago

6 0

The answer is A. When the the temperature increases the kinetic energy increases

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How many photons are absorbed during the dental x-ray?

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<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
E = 7.22182 x 10^-15 J
To convert to eV divide by 1.6 x 10^-19
E = 7.22182 x 10^-15/1.6 x 10^-19 eV
E =45.36 x 10^3 eV
Th energy, E, of a single x-ray photon in eV is = 45.36keV.

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n = 3.85 x 10^-6/7.22182 x 10^-15
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Rainwater draining from a neighborhood street initially travels at 4 ft/s through a pipe with a cross-sectional area of 15.7 ft2

Fudgin [204]

Answer:

The velocity is $v_2 = 0.96 \ ft/s$

Explanation:

From the question we are told that

The initial speed is $v_1 = 4 \ ft/s$

The cross -sectional area of the first pipe is $A_1 = 15.7 \ ft$

The cross -sectional area of the second pipe is $A_2 = 65.4 \ ft$

Generally from continuity equation we have that

$A_1 * v_1 = A_2 * v_2$

So

$v_2 = \frac{A_1 * v_1 }{A_2 }$

=> $v_2 = \frac{15.7 * 4 }{65.4 }$

=> $v_2 = 0.96 \ ft/s$

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3 years ago

Dinitrogen oxide has?

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2 years ago

A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above

Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed $v_{1}= 2\ m/s$

Speed $v_{2}=7\ m/s$

Pressure in main pipe $P_{1}=2\times10^{5}\ Pa$

(I). We need to calculate the diameter

Using equation of continuity

$Av_{1}=Av_{2}$

$\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}$

$(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7$

$d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}$

$d_{2}=5.345\ cm$

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

$P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})$

$P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)$

$P_{2}=1.28500\times10^{5}\ Pa$

(III). If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}$

$P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})$

Here, $h_{3}=0$

Put the value in the equation

$P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17$

$P_{3}=3.5400\times10^{5}\ Pa$

Hence, This is required solution.

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2 years ago