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MatroZZZ [7]
3 years ago
6

If we have a one-electron atom, and the electron can occupy any of 5 different energy levels, and transitions are possible betwe

en any level and any other, how many lines might appear in that atom’s spectrum?
Physics
1 answer:
xeze [42]3 years ago
7 0

Answer:

10

Explanation:

 Lets say that a,b,c,d,e are the five allowed energy states in order of decreasing energy. Then the number of possible different spectral lines comes from the electron dropping from a high state to a lower state. So, they will do so in following ways:

a - b

a - c

a - d

a- e

b - c

b - d

b- e

c - d

c- e

d- e

Ans- Ten different possible energy jumps giving six different colors or lines in the spectrum.

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What is an easy way to balance chemical equations?
Lorico [155]

write out what you have on both sides, then just use basic multiplication to try and even out both sides. I can help if you need me to balance some for you!!

8 0
3 years ago
What is the difference between the initial position and the final position of an object?
Grace [21]
The initial is where you are starting and the final postion is where the object ends up
5 0
3 years ago
An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode
sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell v_{0} in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

                v_{x}=v_{0} \times \cos \theta

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of v_{x} and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

           v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}

           v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}

For motion with constant acceleration, we know

            s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}

Along the horizontal, x-axis, we might write this as

            x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}

Measuring distances relative to the firing point means

               x_{0}=0

we know that,

              a_{x}=0

or,

             v_{x}=v_{x 0}=\text { constant }

By applying the values, we get,

           x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}

The acceleration of gravity is vertically downward and is g=-9.8 \mathrm{m} / \mathrm{s}^{2} , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

           y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}

we know, y_{0}=0 and a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}, so,

          y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)

                 y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0
3 years ago
Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them
denpristay [2]

Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

and r is the distance between them.

F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²

F = 1.33×10⁻¹⁰ N

3 0
3 years ago
Why might you want to use a single fixed pulley?
hjlf

A single fixed pulley can be used to raise or lower lightweight objects.

Option b

<u>Explanation:</u>

A pulley is a simple machine tool which is used to make lifting or lowering tasks easy.  A single fixed pulley is a system involving only one pulley fixed on a constant rigid support with a rope wrapped around the wheel. Such a system can be used only to change the direction of  applied force in raising or lowering small, lightweight objects which need minimal work force.

A single fixed pulley system helps only in redirecting the applied force direction by using a rope and wheel assembly. The work done in such a case remains the same and hence it is not preferred to use it in lifting heavy objects. Neither is the required force reduced in case of a single fixed pulley system. A movable pulley helps in achieving (A) and (C).

4 0
3 years ago
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