Answer:
AM has longer wavelength
Explanation:
The relation between the wavelength and teh frequency is given by
v = f x λ
Where, f is the frequency and λ be the wavelength.
It shows that the wavelength is inversely proportional to the frequency.
So, higher the frequency, smaller be the wavelength.
So, FM has high frequency than AM, thus, FM has lower wavelength as compared to AM.
Answer:
Momentum of block B after collision =
Explanation:
Given
Before collision:
Momentum of block A =
= 
Momentum of block B =
= 
After collision:
Momentum of block A =
= 
Applying law of conservation of momentum to find momentum of block B after collision
.

Plugging in the given values and simplifying.


Adding 200 to both sides.


∴ 
Momentum of block B after collision =
Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of
<em>F</em> = <em>m a</em> → <em>a</em> = <em>F </em>/ <em>m</em>
so that the cart's final speed is
<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>
<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>
<em />
If we force is halved, so is the accleration:
<em>a</em> = <em>F</em> / <em>m</em> → <em>a</em>/2 = <em>F</em> / (2<em>m</em>)
So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give
(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>