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3 years ago

I need help on this

Physics

1 answer:

Alecsey [184]3 years ago

5 0

Answer:

Tectonic Plate Boundary

Explanation:

As Florida is not close to the edge of tectonic plate.

Hope it helps!

You might be interested in

Can someone pls match these

Lostsunrise [7]

Temperature - The measure of the average kinetic energy of a substance

Internal Energy - Total potential and kinetic energies of the particles of a substance

Heat - Thermal energy that flows from one substance to another

Thermal Energy - Part of internal energy that can be transferred

6 0

3 years ago

Following are the different layers of the sun’s atmosphere. Rank them based on the order in which a probe would encounter them w

blsea [12.9K]

Answer:

Going from earth to the sun a probe would encounter the next layers in order:

Corona
Transition Region
Chromosphere
Photosphere
Convection Zone
Radiative Zone
Core

A brief description of them:

Corona is the outermost layer and it cannot be seen with the naked eye, is starts at about 2100 km from the surface of the sun and it has no limit defined.

Transition Region is between the corona and the chromosphere, it has an extension of about 100km

The chromosphere is between 400 km from the surface of the sun to 2100 km. In this layer the further you get away from the sun it gets hotter.

The photosphere is the surface of the sun, the part that we can see, and extends from the surface to 400km.

The convection zone is where convection happens, hot gas rises, cools and rises again.

Radiative Zone is where the photons try to rise to move to higher layers.

The core of the Sun is where nuclear fusion occurs due to the very high temperatures.

6 0

3 years ago

If the gas inside the flask in the above exercise is cooled so that its pressure is reduced to a value of 715.7 torr, what will

evablogger [386]

Answer:

(a) 49.0 mm

Explanation:

#First we need to calculate the height of the mercury arm:

Reduced pressure is 715.7torr, Initial Pressure is 797.3torr, Required pressure is:

$P_g_a_s=P_a_t_m+P_h\\$

The difference in the two arms will give the pressure difference between the gas placed in the flask attached to an open-end mercury manometer.

$P_g_a_s=P_a_t_m+P_h\\797.3=P_a_t_m+(136.4mm-103.8mm)\\P_a_t_m=764.7mm\\\\h_t_o_t_a_l=136.4mm+103.8mm=240.2mm\\\\P_g_a_s=P_a_t_m+\bigtriangleup h\\\bigtriangleup h=P_g_a_s-P_a_t_m\\\\\bigtriangleup h=715.7mm-764.7mm\\=-49mm$

Hence the height of the mercury in the arm is 49.00mm

4 0

3 years ago

What is the energy of a mole of photons that have a wavelength of 413 nm? (h = 6.626 × 10⁻³⁴ J • s and c = 3.00 × 10⁸ m/s)

Triss [41]

Answer:

Energy of one mole of photon will be $2.89\times 10^5J$

Explanation:

We have given wavelength of photon $\lambda =413nm=413\times 10^{-9}m$

Velocity of light is given $c=3\times 10^8m/sec$

Plank's constant $h=6.626\times 10^{-34}Js$

Energy of the photon is given by $E=\frac{hc}{\lambda }=\frac{6.626\times 10^{-34}\times 3\times 10^8}{413\times 10^{-9}}=0.048\times 10^{-17}J$

We have to find the energy of one mole of photon

One mole of photon is equal to $6.023\times 10^{23}photon$

So energy of one mole of photon will be equal to $6.023\times 10^{23}\times 0.048\times 10^{-17}=2.89\times 10^5J$

So energy of one mole of photon will be $2.89\times 10^5J$

6 0

3 years ago

An armada of spaceships that is 1.60 ly long (in its rest frame) moves with speed 0.800c relative to the ground station in frame

miskamm [114]

Answer:

(a). The trip take a time is 3.61 year.

(b). The trip take a time is 3.96 year.

(c). The trip take a time is 8.73 year.

Explanation:

Given that,

Length = 1.60 ly

Speed of spaceships= 0.800c

Speed of messenger = 0.910 c

(a). We need to calculate the velocity of armada

Using formula of velocity

$v'=\dfrac{v-v_{m}}{1-\dfrac{vv_{m}}{c^2}}$

Put the value into the formula

$v'=\dfrac{0.800c-0.910c}{1-\dfrac{0.800\times0.910 c^2}{c^2}}$

$v'=\dfrac{-0.11c}{1-0.800\times0.910}$

$v'=-0.404c$

We need to calculate the length

Using formula of length

$l=l'\sqrt{1-\dfrac{v'^2}{c^2}$

Put the value into the formula

$l=1.60\sqrt{1-(-0.404)^2}$

$l=1.46\ ly$

We need to calculate the length of the trip

Using formula of time

$t=\dfrac{l}{|v'|}$

Put the value into the formula

$t=\dfrac{1.46}{0.404}$

$t=3.61\ year$

(b). If the armada's rest frame

We need to calculate the length

Using formula of length

$l=l'\sqrt{1-\dfrac{v'^2}{c^2}$

Put the value into the formula

$l=1.60\sqrt{1-(0)^2}$

$l=1.60\ ly$

Using formula of time

$t=\dfrac{l}{|v'|}$

Put the value into the formula

$t=\dfrac{1.60}{0.404}$

$t=3.96\ year$

(c). If an observer in frame S

We need to calculate the length

Using formula of length

$l=l'\sqrt{1-\dfrac{v'^2}{c^2}$

Put the value into the formula

$l=1.60\sqrt{1-(0.800)^2}$

$l=0.96\ ly$

We need to calculate the time

$t=\dfrac{0.96}{0.910-0.800}$

$t=8.73\ year$

Hence, (a). The trip take a time is 3.61 year.

(b). The trip take a time is 3.96 year.

(c). The trip take a time is 8.73 year.

8 0

3 years ago