We use the Rydberg Equation for this which is expressed as:
<span>1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
</span>
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<span>. Thus,
</span>
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.
The coin's acceleration is <u>0.37 m/s²</u>
Acceleration is the rate of change of the velocity of an item with appreciation to time. Accelerations are vector portions. The orientation of an item's acceleration is given by the orientation of the net pressure appearing on that object.
<u>Calculation:-</u>
<u />
V² = U -2aS
a = U/2S
= 2/2×2.7
= <u>0.37 m/s²</u>
Acceleration is the charge at which velocity modifications with time, in terms of each speed and route. A factor or an object moving in a straight line is accelerated if it quickens or slows down. movement on a circle is extended despite the fact that the rate is consistent because the course is continually changing.
Learn more about acceleration here:- brainly.com/question/29110429
#SPJ9
The height of the object will be -5.19 cm
A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.
Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall
So let,
v = Image distance from the mirror = -33.5 cm
u = object distance from the mirror (concave) = 24 cm
hi = Image height = 7.25 cm
h = height of the object = ?
Using below formula to find height of the object
-v/u = hi/h
Putting all value in the formula we get
-(-33.5)/(-24) = 7.25/h
h = -5.19 cm
Therefore the height of the object will be -5.19 cm
Learn more about Concave mirror here:
brainly.com/question/3727024
#SPJ10
Incomplete question as there is so much information is missing.The complete question is here
A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?
Answer:
Distance traveled=240 m
Explanation:
Given data
Initial velocity of car v₀=0 m/s
Final velocity of car vf=24 m/s
Distance traveled by car S=120 m
To find
Distance does the traffic travel
Solution
To find the distance first we need to find time, for time first we need acceleration
So
As we find acceleration.Now we need to find time
So
Now for distance
So
