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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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Me ajudem, Por favor!!!!!!
zzz [600]

Answer:

a)  a=4\,\frac{m}{s^2}

b)  V(t)=4\,t\,+3

c)  V(1)=7 \,\frac{m}{s} \\

d)  Displacement = 22 m

e)  Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}

Therefore,  acceleration is a=4\,\frac{m}{s^2}

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

y=m\,x+b\\V(t)=4\,t\,+3

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = \frac{(7+15)\,2}{2} = 22\,\,m

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = \frac{22}{2} = 11\, \,\frac{m}{s}

3 0
4 years ago
the force that is exerted when a shopping cart is pushed. the forces that causes a metal ball to move toward a magnet
alexandr402 [8]

Answer:

The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.

The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.

Explanation:

Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.

Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.

8 0
3 years ago
Read 2 more answers
What is the net force on an object that has a force pushing downward at 25N and a forcing pushing upward at 10N?
shutvik [7]

Answer:

15N downwards

Explanation:

Net force is the sum of all forces acting on an object.

Taking upwards as positive and downwards as negative,

Net Force = 10N + (-25N)

= -15N

Hence <u>Net Force is 15N downwards</u>.

7 0
4 years ago
Read 2 more answers
Give reasons:<br>A freely suspended magnet shows<br>N-S poles at rest.​
aniked [119]

Answer:

A freely suspended magnet comes to rest along the N-S direction.

Explanation:

-Irrespective of direction of suspension, the bar eventually comes to rest at earth's N-S direction.

-Bar end that points to Earth's north is called the <em>North Pole</em>. Bar end that point's to earth's south is called the <em>South pole</em>.

-The reason for this default positioning is called the Directive Property. This is because earth acts as an external magnetic field.

7 0
4 years ago
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
4 years ago
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