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3 years ago

The wavelengths of light emitted by an excited atom are determined by the:

Physics

2 answers:

Natalka [10]3 years ago

7 0

Difference in energy between the exicted states and ground states

Murrr4er [49]3 years ago

3 0

The best and most correct answer among the choices provided by your question is the first choice or letter A.

<span>The wavelengths of light emitted by an excited atom are determined by the: differences in energy between the excited states and ground state.
</span>

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How are energy sources managed?<br><br> Could you link a site with a good answer?

Arturiano [62]

Answer: When it comes to energy saving, energy management is the process of monitoring, controlling, and conserving energy in a building or organization. Typically this involves the following steps: Metering your energy consumption and collecting the data.

Explanation:

8 0

2 years ago

Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find

True [87]

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion $s=ut+\frac{1}{2} at^2$ , where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = $-9.8m/s^2$ .

$-20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0$

t = 6.53 seconds or t = -0.63 seconds

So time = 6.53 seconds.

Considering the horizontal motion of arrow

u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = $0m/s^2$

$s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m$

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.

7 0

3 years ago

A mechanical device requires 500j of work to do 200j of work in lifting a box. What is the efficiency of the device?

olga55 [171]

40% the formula is work output/work input*100 so
200/500=.4*100=40

5 0

3 years ago

A 320-kg satellite experiences a gravitational force of 800 N. What is the radius of the satellite’s orbit? What is its altitude

Westkost [7]

Answer:

Radius of orbit = 3.992 × $10^{7}$ m

Altitude of Satellite =33541.9× m

Explanation:

Formula for gravitational force for a satellite of mass m moving in an orbit of radius r around a planet of mass M is given by;

$F= G\frac{m M}{r^{2} }$

Where G = Gravitational constant = 6.67408 × 10-11 $\frac{m^{3} }{Kg sec^{2} }$

We are given

F= 800 N

m = 320 Kg

M = 5.972 × $10^{24}$ Kg

G = 6.67408 × 10-11 $\frac{m^{3} }{Kg sec^{2} }$

We have to find radius r =?

putting values in formula;

==> 800 =6.67408 × $10^{-11}$ × 320 × 5.972 × $10^{24}$ / $r^{2}$

==> 800= 39.8576 × $10^{13}$ × 320 / $r^{2}$

==> 800 = 12754.43 × $10^{13}$ / $r^{2}$

==> $r^{2}$ = 12754.43 × $10^{13}$ /800

==> $r^{2}$ =15.94 × $10^{13}$

==> r = 3.992 × $10^{7}$ m

==> r = 39920× $10^{3}$ m

This is the distance of satellite from center of earth. To find altitude we need distance from surface of earth. So we will subtract radius of earth from this number to find altitude.

Radius of earth =6378.1 km = 6378.1 × $10^{3}$ m

Altitude = 39920× $10^{3}$ - 6378.1 ×

3 0

3 years ago

The strength of the electric field 0.5 m from a 6 µc charge is n/c. (use k = 8.99 × 109 n•meters squared per coulomb squared and

Kruka [31]

The strength of the electric field will be 215760 NC⁻¹. The concept of the electric field strength is used in the problem.

<h3>What is electric file strength?</h3>

The electric field strength is defined as the ratio of electric force and charge. Its unit is NC⁻¹.

The given data in the problem is;

E is the Electric Field Strength

k is the Colomb's constant = 8.99 x 10⁹ N.m²/C²

q is the magnitude of charge = 6 μC = 6 x 10⁻⁶ C

r is the distance = 0.5 m

The electric field strength is given by the formula;

$\rm E = \frac{Kq}{R^2} \\\\ \rm E = \frac{8 .99 \times 10^9 6 \times 10^6 }{(0.5)^2} \\\\ E=215760 \ NC^{-1}$

Hence the strength of the electric field will be 215760 NC⁻¹.

To learn more about the electric field strength refer to the link;

brainly.com/question/4264413

6 0

2 years ago