If al snow and glaiers melted, would the sea level rise or fall

A line from the Brackett series of hydrogen has a wavelength of 1945nm (or 1.0979x10^7m). From which state did the electron orig

Zarrin [17]

We use the Rydberg Equation for this which is expressed as:

1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]

where lambda is the wavelength, where n represents the final and initial states. Brackett series means that the initial orbit that electron was there is 4 and R is equal to 1.0979x10^7m. Thus,

1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
1/1.0979x10^7m = 1.0979x10^7m [ 1/(n2)^2 - 1/(4)^2]

Solving for n2, we obtain n=1.

5 0

3 years ago

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

tatuchka [14]

Answer: The final temperature of the solution is $80.14^oC$

Explanation:

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

4 0

3 years ago

Both objects A and D represent fixed, negatively-charged particles of equal magnitude and Object B represents a fixed, positivel

Kipish [7]

Answer:

Option A = 1.

Explanation:

So, in order to solve this question we are given the Important infomation or data or parameters in the question above as;

(1). First, Both objects A and D represent fixed.

(2). Both objects A and D are negatively-charged particles of equal magnitude.

(3). "Object B represents a fixed, positively-charged particle (equal, but opposite charge from A and D)."

(4). "Object C shows a moving, positively-charged particle."

So, our mission is to determine the arrow that would correctly show the force of attraction or repulsion on object C caused by the other two objects.

We can do that by drawing out the forces of attraction and the resultants. Therefore, CHECK THE ATTACHED FILE/PICTURE FOR THE DRAWINGS.

The forces of attraction due to objects A and B on on object C will be towards themselves. Hence, the resultant is ONE(1).

6 0

3 years ago

Pertaining to simple machines and levers what changes when the fulcrum position is modified

ladessa [460]

Explanation :

Simple machines makes our work easier. Lever is one of the simple machine which consists of rigid rod that is pivoted at a fixed support called as Fulcrum.

There are three classes of lever.

Class 1 : In this type of class, fulcrum is placed in between effort and load. Hence the movement of load is in reverse direction of the movement of effort. (fig 1)

Class 2 : In this type of, the load is between the effort and the fulcrum. Hence, the movement of load is in same direction as that of the effort. (fig 2)

Class 3 : In this type of lever the effort between the load and the fulcrum. Hence, both the effort and load are in same direction. (fig 3)

Hence, when the position of fulcrum is modified the effort force changes.

6 0

3 years ago

Explain briefly why the intensity reflected off the back surface of the film (i.e., the right surface, where there is a liquid-t

Katarina [22]

Explanation:

Taking the incident light to be traveling in the + x-direction so that it is at normal incidence to the left side of the film(referred to as the "Front side"). This means the beam transmitted into the liquid is essentially as strong as the incident beam.

Almost all the light that is reflected off the back surface will get through the front surface. (But only 2.78% gets re-reflected off the the front surface back to the right) this means that there are two beams reflected to the - x-direction, one from the front surface and one from the back, and these beams are of almost equal intensity.

Hope this is helpful. Thanks

7 0

3 years ago