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Ivanshal [37]
4 years ago
13

6. The table shows the first five energy levels for mercury. A mercury atom makes a transition that emits a photon with a freque

ncy of 4.91 × 1014 Hz. What was the transition? Use GUESS Method!!
Energy levels of mercury:
n energy level eV
1 -10.38
2 -5.74
3 -5.52
4 -4.95
5 -3.71

Physics
1 answer:
bezimeni [28]4 years ago
7 0
Now,
E= 2.03 eV
Also, 
Ef – Ei = E2 – E5 
= -5.74 eV – (–3.71 eV)
= –2.03 eV.
From the conservation of energy, we know that the atom lost energy when that atom emitted the photon. Thus, the transition was from E5 to E2.

The energy levels for an atom are quantized.

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A 0.91 kg ball and a 2 kg ball are connected by a 0.95 m long rigid, massless rod. The rod is rotating clockwise about its cente
Sloan [31]

Answer: 0.235

Explanation:

Given data.

M1 = 0.91 Kg

M2 = 2kg

L = 0.95m

N = 35rpm

t = 10.44secs

Solution:

• we first solve for the center of gravity

•we calculate our moment of inertia

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6 0
3 years ago
A train 400 m long is moving on a straight track with a speed of 81.4 km/h. The engineer applies the brakes at a crossing, and l
Luda [366]

Answer:

86.5m

Explanation:

first convert km/h

then

81.4*1000/60*60=22.6

17.6*1000/60*60=4.89

then, x1/t1=x2/t2

we get

x2=400*4.89/22.6=86.5//

3 0
3 years ago
What electrons are in “motion”, what do you have?
Sergeeva-Olga [200]

Answer:

When a positive charged object is placed near a conductor electrons are attracted the the object. ... When electric voltage is applied, an electric field within the metal triggers the movement of the electrons, making them shift from one end to another end of the conductor. Electrons will move toward the positive side. As you know, electrons are always moving. They spin very quickly around the nucleus of an atom. As the electrons zip around, they can move in any direction, as long as they stay in their shell.

7 0
4 years ago
The answer and how to do it
Arturiano [62]
Current = charge per second
2 Coulombs per second = 2 Amperes

Potential difference = (current)x(resistance) in volts.

That's (2 Amperes) x (2 ohms).

That's how to do it.
I think you can find the answer now.
8 0
4 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
3 years ago
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