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3 years ago

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Which is an SI base unit that makes up part of the unit of force?

1 answer:

Snezhnost [94]3 years ago

5 0

Answer:

Which is an SI base unit that makes up part of the unit of force?

1.candela

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In which terrestrial biome would you expect to rake leaves? Question 3 options: temperate deciduous forest taiga (coniferous for

Free_Kalibri [48]

Temperate Deciduous Forest.

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3 years ago

A sample of an ideal gas is slowly compressed to one-half its original volume with no change in pressure. If the original root-m

Zigmanuir [339]

Answer:

$V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}$

So then the final answer on this case would be:

$\frac{V_{rms_i}}{\sqrt{2}}$

Explanation:

From the kinetic theory model of gases we know that the velocity rms (speed of gas molecules) is given by:

$V_{rms}= \sqrt{\frac{3RT}{M}}$ (1)

Where V represent the velocity

R the constant for ideal gases

T the temperature

M the molecular weight of the gas

We also know from the ideal gas law that $PV= nRT$

If we solve for T we got: $T = \frac{PV}{nR}$

For the initial state we can replace T into the equation (1) and we got:

$V_{rms_i}= \sqrt{\frac{3R (\frac{PV}{nR})}{M}} = \sqrt{\frac{3PV}{M}}$

For the final state we know that : $V_f = \frac{V}{2}$ And the pressure not change , so then the final velocity would be:

$V_{rms_f}= \sqrt{\frac{3R (\frac{P(V/2)}{nR})}{M}} = \sqrt{\frac{3P(V/2)}{M}}$

$V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}$

So then the final answer on this case would be:

$\frac{V_{rms_i}}{\sqrt{2}}$

6 0

3 years ago

A 10-kg piece of aluminum sits at the bottom of a lake, right next to a 10-kg piece of lead, which is much denser than aluminum.

Lelechka [254]

I think that the answer is B

5 0

3 years ago

Read 2 more answers

A light beam with wavelength 500 nm is reflected constructively from a thin layer of oil having index of refraction 1.25. The oi

kherson [118]

Answer:

t = 200 nm

Explanation:

It is given that,

Wavelength of a light beam is 500 nm

It is reflected constructively from a thin layer of oil having index of refraction 1.25.

The oil floats on the top of water of index of refraction 1.33.

We need to find the minimum thickness of the layer of oil. `

For constructive interference,

$\dfrac{2nt}{\lambda}=m$

m is 1 for minimum thickness

n is refractive index of oil

t is thickness of layer of oil.

So,

$t=\dfrac{m\lambda}{2n}\\\\t=\dfrac{1\times 500\times 10^{-9}}{2\times 1.25}\\\\t=2\times 10^{-7}\ m\\\\t=200\ nm$

So, the minimum thickness of the layer of oil is 200 nm.

5 0

3 years ago

A lightbulb manufacturer makes bulbs with different "color temperatures," meaning that the spectrum of light they emit is simila

Oksana_A [137]

Answer:

The bulb with higher temperature(4000 K) will be brighter

Explanation:

From the question we are told that

The color temperature for first bulb is $T_1 = 2000K$

The color temperature for second bulb is $T_2 = 4000K$

Generally the emission power of black body radiation is mathematically represented as

$E = \sigma T^4$

Where $\sigma$ is the Stefan-Boltzmann constant with a value $\sigma = 5.67 * 10^{-8} W m^{-2} K^{-4.}$

Now for $T_1 = 2000K$

$E_1 = 5.67*10^{-8} * (2000)^4$

$E_1 = 907.2 \ W/m^2$

At $T_2 = 4000K$

$E_2 = 5.67*10^{-8} * 4000$

$E_2 = 14515.2 \ KW/m^2$

Looking at the result we got we see that the emission power for the higher temperature bulb is higher, this means that its power to emit in the visible spectrum range would be higher

So the bulb with higher temperature will be brighter

7 0

3 years ago