Answer:
number 4
Explanation:
The reflection of light happens when the light bounces off the reflecting surface. That is described by the last (bottom) schematics.
Therefore, select answer number 4.
Answer:
x = 0.396 m
Explanation:
The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring
Data the putty has a mass m1 and velocity vo1, the block has a mass m2
. t's start using the moment to find the system speed.
Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash
p₀ = m1 v₀₁
Moment after shock
= (m1 + m2) 
p₀ =
m1 v₀₁ = (m1 + m2) 
= v₀₁ m1 / (m1 + m2)
= 4.4 600 / (600 + 500)
= 2.4 m / s
With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring
Before compressing the spring
Em₀ = K = ½ (m1 + m2)
²
After compressing the spring
= Ke = ½ k x²
As there is no rubbing the energy is conserved
Em₀ = 
½ (m1 + m2)
² = = ½ k x²
x =
√ (k / (m1 + m2))
x = 2.4 √ (11/3000)
x = 0.396 m
Answer:
The value is 
Explanation:
From the question we are told that
The focal length of the objective is 
The focal length of the eyepiece is 
The tube length is 
Generally the magnitude of the overall magnification is mathematically represented as

Where
is the objective magnification which is mathematically represented as

=> 
=> 
is the eyepiece magnification which is mathematically evaluated as



So


Explanation:
The frequency of radio waves is 1.667 GHz
One portion of the same wave front travels 1.260 mm farther than the other before the two signals are combined.
There are two conditions for interference either constructive or destructive.
For constructive interference , the path difference is n times of wavelength and for destructive interference, the path difference is (n+1/2) times of wavelength
We can find wavelength in this case as follows :

If we divide path difference by wavelength,

It means that the path difference is 7 times of the wavelength. it means the two waves combine constructively and the value of m for the path difference between the two signals is 7.