Answer:
a) λ = 189.43 10⁻⁹ m b) λ = 269.19 10⁻⁹ m
Explanation:
The diffraction network is described by the expression
d sin θ= m λ
Where m corresponds to the diffraction order
Let's use trigonometry to find the breast
tan θ = y / L
The diffraction spectrum is measured at very small angles, therefore
tan θ = sin θ / cos θ = sin θ
We replace
d y / L = m λ
Let's place in the first order m = 1
Let's look for the separation of the lines (d)
d = λ L / y
d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²
d = 332.33 10⁻⁹ m
Now we can look for the wavelength of the other line
λ = d y / L
λ = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²
λ = 189.43 10⁻⁹ m
Part B
The compound wavelength B
λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²
λ = 269.19 10⁻⁹ m
Answer:
C. At a particular instant
Explanation:
Speed is the defined as the ratio between the distance covered by an object and the time taken:
![v=\frac{d}{t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bt%7D)
where d is the distance and t the time.
However, there are two possible measurements of speed:
- Average speed: this is the speed measured over a non-zero time interval (for example: a car moving 100 metres in 5 seconds; its average speed is
![v=\frac{100 m}{5 s}=20 m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B100%20m%7D%7B5%20s%7D%3D20%20m%2Fs)
- Instantaneous speed: this is the speed of an object measured at a particular instant in time, so for a time interval that tends to zero. So, in the previous example, the average speed is 20 m/s but the instantaneous speed of the car at various instants of time can be different from that value.
Answer: 20.4 miles
Explanation:
Here we need to use the equation:
Velocity = Distance/Time.
Initially we have that he can travel 30 miles in 2 hours, so the velocity is:
V = 30mi/2h = 15mph
Now, we reduce the velocity by 3 mph, so the new velocity is 15mph - 3 mph = 12mph.
Now we want to know the distance traveled in 1.7 hours with this velocity, this is.
Velocity*Time = Distance
12mi/h*1.7h = 20.4 miles
Answer:
force-strength,power or energy as an attribute of motion, movement or action. Example: Frictional force.
Answer:
The image distance is 17.56 cm
Explanation:
We have,
Height of light bulb is 3 cm.
The light bulb is placed at a distance of 50 cm. It means object distance is, u =-50 cm
Focal length of the lens, f = +13 cm
Let v is distance between image and the lens. Using lens formula :
![\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{13}+\dfrac{1}{(-50)}\\\\v=17.56\ cm](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7Bv%7D-%5Cdfrac%7B1%7D%7Bu%7D%5C%5C%5C%5C%5Cdfrac%7B1%7D%7Bv%7D%3D%5Cdfrac%7B1%7D%7Bf%7D%2B%5Cdfrac%7B1%7D%7Bu%7D%5C%5C%5C%5C%5Cdfrac%7B1%7D%7Bv%7D%3D%5Cdfrac%7B1%7D%7B13%7D%2B%5Cdfrac%7B1%7D%7B%28-50%29%7D%5C%5C%5C%5Cv%3D17.56%5C%20cm)
So, the image distance is 17.56 cm.