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3 years ago

The force that pulls planets torwards the sun is called

Physics

1 answer:

Studentka2010 [4]3 years ago

7 0

Answer:

gravity

Explanation:

Gravity pulls the planets toward the Sun. Gravity pulls the moon toward Earth. Gravity pulls us toward the Earth. Gravity is a force. Inertia.

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When the hydrogen atom makes the transition from the n=2 to the n=1 energy level, it emits a photon. This photon can be absorbed

bazaltina [42]

Answer:

$n_{fn}$ = 4

Explanation:

To solve this exercise we will use Bohr's atomic model

$E_{n}$ = - 13.606 / n² [eV]

The transition from level n = 2 to level n = 1 is valid

$E_{21}$ = - 13.606 [¼ -1/1]

$E_{21}$ = 10.2045 eV

Bohr's model for atoms with only one electron is

$E_{n}$ = -13.606 Z² / n²

Where Z is the atomic number of the atom.

In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches

ΔE = -13.606 [4 / $n_{fn}$ ² - 4/4]

4 / $n_{fn}$ ² = -ΔE / 13.606 + 1

4 / $n_{fn}$ ² = -10.2045 / 13.606 +1 = -0.75 +1

4 / $n_{fn}$ ² = 0.25

$n_{fn}$ = √ 4 / 0.25

$n_{fn}$ = 4

8 0

3 years ago

A fluid is flowing through a circulat tube at 0.4 kg/s. Tube inner surface is smooth with a diameter 0.014 m. Fluid density is 9

ElenaW [278]

Answer:

The convection coefficient is $15456.48\ W/m^{2}K$

Solution:

Mass flow rate, $\dot{m} = 0.4\ kg$

Inner diameter of the tube, d = 0.014 m

Fluid density, $\rho_{f} = 990\ kg/m^{3}$

Specific Heat, C = 3845 J/K

Thermal Conductivity, K = 0.74

Prandtl Number, $P_{r} = 8.6$

Heat flux, $\dot{q} = 71,297\ W/m^{2}$

Viscosity, $\mu = 0.00079\ Ns/m^{2}$

Now,

To calculate the convection heat coefficient, h:

Determine the cross sectional area of the circular tube:

$A_{c} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}\times (0.014)^{2} = 1.54\time 10^{- 4}\ m^{2}$

Determine the velocity of the fluid inside the tube by mass flow rate:

$\dot{m} = \rho_{f}A_{c}v$

$0.4 = 990\times 1.54\time 10^{- 4}v$

v = 2.624 m/s

Determine the Reynold's Number, $R_{e}$ :

$R_{e} = \frac{\rho_{f}dv}{\mu}$

$R_{e} = \frac{990\times 0.014\times 2.624}{0.00079} = 46036.253$

Thus it is clear that $R_{e}$ > 10,000 hence flow is turbulent.

Now,

Determine the Nusselt Number:

$N_{u} = 0.023R_{e}^{0.8}P_{r}^{0.4}$

$N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42$

Also,

$N_{u} = \frac{dh}{K}$

where

h = convection coefficient

Now,

$292.42 = \frac{0.014\times h}{0.74}$

$h = 15456.48\ W/m^{2}K$

7 0

3 years ago

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called Electrostatic Force.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0

3 years ago

PLEASE PLEASE HELP AND PUT A REAL ANSWER ;-;. ALSO WILL GIVE BRAINLEST!!

uysha [10]

Answer:

Troposphere

High-pressure areas form due to downward motion through the troposphere, the atmospheric layer where weather occurs.

3 0

2 years ago

Read 2 more answers

A concise definition of pair production

kogti [31]

Pair production is a direct conversion of radiant energy to matter. It is one of the principal ways in which high-energy gamma rays are absorbed in matter.

6 0

3 years ago

The force that pulls planets torwards the sun is called​

The force that pulls planets torwards the sun is called