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3 years ago

What is released when an excited electron returns to a lower energy state?

Physics

2 answers:

Marysya12 [62]3 years ago

8 0

Answer:

A proton

Explanation:

oee [108]3 years ago

8 0

Answer:

D. Electromagnetic radiation

Explanation:

...

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Which of the following is considered a calm area close to the equator resulting in little to no wind?

solong [7]

<h2>Right answer: Doldrums</h2>

These are also called zones of equatorial calm and it is due a climatic phenomenon that is placed near the Earth equator, attributed to the soft winds, that are called calm winds as well; accompanied by systems of abundant rains and heat.

In this area periods of great calm occur when the winds virtually disappear completely, trapping the sailing ships for long periods (days or weeks). This is why the term <em>doldrum</em> became popular as a colloquial expression in the eighteenth century, to refer to "<em>the caprice of the wind that slows down the navigation to sail". </em>

The zone is located in the place where two trade winds meet, this means the trade winds of the northern hemisphere <u>converge</u> with those of the southern hemisphere, that is why this region is related to the <u>intertropical convergence zone</u>.

5 0

4 years ago

I give a ball a push on an acclivity. The "start velocity" is on 7m/s. The time it took the ball to get back to me was 10 second

Degger [83]

Answer:

7÷10

Explanation:

initial velocity=7m/s

final velocity=0m/s

3 0

4 years ago

Read 2 more answers

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

Svetllana [295]

Answer:

$P_G = 14.03 \ psig$

$h_m = 0.148 \ m$

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is $P_{m} = 15.5 \ psig = 15.5 * 6894.76 = 106868.78 \ N/m^2$

The level of mercury is $h = 950 \ mm = 0.950 \ m$

The drop in the mercury level at the visible arm is $d = 39.0 = 0.039 \ m$

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

$P_g = P_m = g * \delta h * \rho$

Here $\rho$ is the density of mercury with value $\rho = 13.6 *10^{3} kg/m^3$

and $\delta h$ is the difference in the level of gas in arm one and two

$\delta h = \frac{106868.78}{ 13.6 *10^{3} * 9.8 }$

$\delta h = 0.802 \ m$

Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

$h_m = 0.950 - 0.802$

=> $h_m = 0.148 \ m$

Generally from manometry principle we have that

$P_G + \rho * g * d - \rho * g * [h - (h_m + d)] = 0$

Here $P_G$ is the pressure of the gas

$P_G +13.6 *10^{3} * 9.8 * 0.039 - 13.6 *10^{3} * 9.8 * [0.950 - (0.148 + 0.039)] = 0$

$P_G = 9.6724 04 *10^{4} \ N/m^2$

converting to psig

$P_G = \frac{ 9.6724 04 *10^{4} }{6894.76}$

$P_G = 14.03 \ psig$

6 0

3 years ago

Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

3 years ago

Read 2 more answers

A cube of copper was found to have a mass of 630 g. What are the dimensions of the cube? (The density of copper is 8.94 g/cm3.)

raketka [301]

Answer:

Density = mass/volume

Volume of cube = mass/density

630g/8.94 = 70.5cm^3 (3s.f.)

Since it is a cube, its dimensions of Length x breadth x height=volume

Take 70.5 amd cube root it

$\sqrt[3]{70.4697986577)$

=4.13 (3s.f)

hence , it would be 4.13 x 4.13 x 4.13 to find the volume

Hence, the answer would be 4.13cm(3s.f.)

3 0

4 years ago