The magnification of the ornament is 0.25
To calculate the magnification of the ornament, first, we need to find the image distance.
Formula:
- 1/f = u⁻¹+v⁻¹.................... Equation 1
Where:
- f = Focal length of the ornament
- u = image distance
- v = object distance.
make u the subject of the equation
- u = fv/(f+v)................ Equation 2
From the question,
Given:
Substitute these values into equation 2
- u = (12×4)/(12+4)
- u = 48/16
- u = 3 cm.
Finally, to get the magnification of the ornament, we use the formula below.
- M = u/v.................. Equation 3
Where
- M = magnification of the ornament.
Substitute these values above into equation 3
Hence, The magnification of the ornament is 0.25
The relationship between the number of visible spectral lines are identical for atoms .However they have unique wavelengths.
Option B
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Explanation:</u>
A spectrum is a range of frequencies or a range of wavelengths. The photon energy of the emitted photon is equal to the difference between two states. For every atom there are quite many electron transitions and each has a energy difference.
This difference in wavelength causes spectrum .As each element emission spectrum is unique because each atom has different energy and causes uniqueness in the emission spectrum . Hence, due to the difference in energy it emits different wavelengths.
Answer:21.18 m
Explanation:
Given
initial speed u=10 m/s
height of building h=22 m
time taken to complete 22 m
initial vertical velocity =0
Horizontal Distance moved
Answer:
4.08 s
Explanation:
Let the passenger took "t" time to catch the train
so in this case the total distance moved by the train + 5 m = total distance moved by the passenger
so we will have
distance moved by train is given as
also the distance moved by passenger
so we will have
t = 4.08 s
Answer:
Explanation:
The change is as follows
P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process
3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process
In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁
P₁V₁ = n R T₁ , n is no of moles of gas enclosed.
nRT₁ = P₁V₁
Heat added at constant volume = n Cv ( 3T₁ - T₁)
= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)
= 10/3 x nRT₁
= 10/3x P₁V₁
In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁
Heat added at constant pressure in second case
= n Cp ( 15T₁ - 3T₁)
= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)
= 28 x nRT₁
= 28 P₁V₁
