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3 years ago

What does the symbol "E3" represent? ...?

2 answers:

7 0

The symbol "E3" represents the energy of electrons in the third energy level.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

laila [671]3 years ago

4 0

Answer:

E₃ = energy in third energy level.

Explanation:

The energy of an electron is third orbit is given by :

$E=-\dfrac{13.6Z^2}{n^2}$

Where

Z is the atomic number

n is principal quantum number and n = 1,2,3....

i.e. the energy of an electron is inversely proportional to the principal quantum number.

E₁,E₂,E₃.......... shows the energy of an electron in first, second and third energy level respectively.

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Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed

likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let $t$ represents the time in seconds and $v$ the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of $v$ : $\rm 2\;m\cdot s^{-1}$ at $t = 0$ ; Hence the point on the segment: $(0, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .
Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 1$ : $t = 2t + 2 = 4$ ; Hence the point on the segment: $(1, 4)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .

Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 2$ : $t = t + 3 = 5$ ; Hence the point on the segment: $(2, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.
Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 3$ : $t = 5$ ; Hence the point on the segment: $(3, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.
Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .

$t = \rm 3.49\;s$ is in the interval $3 < x \le 4$ . Apply the equation for that interval: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$ .

4 0

3 years ago

A step-down transformer providing electricity for a residential neighbor-hood has exactly 2680 turns in its primary. when the po

Fynjy0 [20]

The turns ratio is equal to the voltage ratio. Let n1 and n2 be the primary and secondary turns. Then
5850V/120V=n1/n2
48.75=2680/n2
n2=2680/48.75
n2=55

6 0

3 years ago

What is the mass of an object travelling at 25 m/s with a kinetic energy of 3775 J?

Arte-miy333 [17]

Answer:

<h2>You can do 6516-3775=2741</h2><h2 /><h2>The difference is 27141 kilometers</h2><h2 /><h2>Did this help?</h2><h2>⇒ Yes or No?</h2>

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3 years ago

When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache

GenaCL600 [577]

Answer:

None of the above forces on air drag on him is equal to his weight

Explanation:

In the velocity-time graph,the gradient of the curve where it is flatten shows the parachutist reaches the terminal velocity when it reaches terminal velocity which means the parachutist reaches constant velocity or speed,indicating that the acceleration of free fall(g) is zero.And according to the resultant force formula weight - air drag= mass*acceleration. so when accelerate is zero,resultant force is zero. And hence the equation will be like this: weight= air drag

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3 years ago

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blagie [28]

Answer:

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3 years ago