Answer:
Main Difference Between Mechanical and Electromagnetic waves
A wave is composed of some kind of disturbance that propagates. We can classify waves into many different types based on their properties. One of the properties of the waves depends on whether they need a medium to propagate or not. The primary difference between electromagnetic and mechanical waves is also based on this property. Mechanical waves need a medium, while electromagnetic waves do not need a medium to propagate. Electromagnetic waves can travel through a vacuum. The other differences between mechanical and electromagnetic waves are given below:
Electromagnetic waves can travel through a vacuum, that is an empty space, whereas mechanical waves cannot. They need a medium to travel such as water or air. Ripples in a pond are an example of mechanical waves whereas electromagnetic waves include light and radio signals, which can travel through the vacuum of space.
Mechanical waves can be classed as elastic waves because their transmission depends on the medium's (water, air etc.) elastic properties.
Electromagnetic waves are caused because of the varying magnetic and electric fields. They are produced by the vibration of the charged particles.
Because of these differences, the speed of each type of wave varies significantly. Electromagnetic waves travel at the speed of light but mechanical waves are far slower.
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Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
Answer:
5.5 m/ sec
Explanation:
Because the inclined surface is frictionless so we can assume that total change of energy is zero
i-e ΔE = 0
Or we can say that difference between final and initial energy is zero i-e
Ef- Ei =0
Where,
Ef= final energy at the top of the ramp= KEf+PEf
Ei= Initial energy at the bottom of the ramp=KEi+PEi
So we have
(KEf+PEf)-(KEi+PEi)=0
==>KEf-KEi+PEf-PEi=0 -------------(1)
KEf = mgh = 200×9.8×h
Where h= Sin 22 = h/d= h/4.1
or
0.375×4.1=h
or h= 1.54 m
So, PEf= 200×9.8×1.54=3018.4 j
and KEf= 1/2 m
= 0.5×200×0=0 j
PEi= mgh = 200×9.8×0=0 j
KEi= 1/2 m
=0.5×200×=100 j
Put these values in eq 1, we get;
0-100 +3018.4-0=0
-100 =-3018.4
==> 
= 30.184
==> Vi = 
Answer: It's hard to say without characterizing the collision. But it will be either A if the collision is totally in-elastic, or B if the collision is totally elastic. It could be anywhere in between for partially elastic collisions.
Explanation:
momentum is conserved, so initial system momentum will be left to right.
The velocity of the center of mass is 50(5) / 550 = 0.4545... m/s
In an elastic collision, the lead ball will move off at twice that speed or 0.91 m/s to the right.
The steel ball will bounce back and move away at 0.91 - 5 = -4.1 m/s . The negative sign indicates the steel ball has reversed course and has negative momentum
In a totally in-elastic collision, both balls would move to the right at 0.45 m/s. The steel ball will still have positive momentum.
