All of the following are categories of individual activities except?

E=<br> (500.0lm)<br> 4 (100)

drek231 [11]

Answer:

didn't understand your question

7 0

3 years ago

Explain what differentiates the Earth’s crust and lithosphere.

Allisa [31]

Answer:

What is the difference between the crust and lithosphere? The crust (whether continental or oceanic) is the thin layer of distinctive chemical composition overlying the ultramafic upper mantle. ... The lithosphere is the rigid outer layer of the Earth required by plate tectonic theory.

Explanation:

Hope this helps!

4 0

3 years ago

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Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p

PSYCHO15rus [73]

Answer:

Energy density will be 14.73 $J/m^3$

Explanation:

We have given capacitance $C=225\mu F=225\times 10^{-6}F$

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm $0.2\times 10^{-3}m$

We know that there is relation between electric field and potential

$E=\frac{V}{d}$ , here E is electric field, V is potential and d is separation between the plates

So $E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C$

Energy density is given by $E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3$

5 0

3 years ago

What variables affect density weight,conductivity,color,volume, or mass

tatiyna

Answer:

Density is affected by volume and mass.

Explanation:

Density is defined as the quantity of mass per unit of volume, or expressed mathematically, d = m/v.

6 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$