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3 years ago

Which item is not considered electromagnetic energy?

1 answer:

5 0

Sound waves are known to be the one that's not considered as a type of electromagnetic energy. As for microwaves and x-rays, they tend to share the same frequencies that can be considered as electromagnetic, and sound waves have a different frequency than them.

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A student wearing frictionless in-line skates on a horizontal surface is pushed, starting from rest, by a friend with a constant

grin007 [14]

Answer:

Physics

Explanation:

Explanation:

We can use the Theorem of Work (W) and Kinetic Energy (K):

W=ΔK=Kf−Ki

it basically tells us that the work done on our system will show up as change in Kinetic Energy:

We know that the initial Kinetic Energy, Ki=12mv2i, is zero (starting from rest) while the final will be equal to 352J; Work will be force time displacement. so we get:

F⋅d=Ff

45d=352

and so:

d=35245=7.8≈8m

8 0

3 years ago

Read 2 more answers

An undamped oscillator has period ro= 1.000 s, but I now add a little damping so that its period changes to r i= 1.001 s.

Firlakuza [10]

This Question has mistakes.Correct question is

An undamped oscillator has period τo=1.000s , but I now add a little damping so that its period changes to τ1=1.001s.

What is the damping factor β? By what factor will the amplitude of oscillation decrease after 10 cycles? Which effect of damping would be more noticeable, the change of period or the decrease of the amplitude?

Answer:

β=0.28 /sec

Amplitude=0.0606

Explanation:

$T_{o}=2\pi /w\\T_{1}=2\pi /\sqrt{w^{2} -\beta^{2} }\\ As\\T_{o}=1 \\T_{1}=1.001\\From T_{o}\\w=2\pi \\So\\T_{1}=w/\sqrt{w^{2} -\beta^{2} }\\\beta =0.00447w\\\beta =0.00447(2\pi )\\\beta =0.28sec^{-1}\\ Amplitude=Ae^{-\beta t}\\ after\\ t=10T_{1}\\so\\Ae^{-\beta(10t_{1}) }=e^{-\beta(10t_{1}) }\\=e^{-\(0.28)(10)(1.001) }\\Amplitude=0.0606$

7 0

4 years ago

A skater of mass 40 kg is carrying a box of mass 5 kg. The skater has a speed of 5 m/s with respect to the floor and is gliding

Yuki888 [10]

For the part a) we need only the momentum of the box and we have the data to find it.

Momentum is given by,

$p=mv$

where clearly, p is the momentum, m the mass of the box and v is the velocity.

Substituting,

$p=(5kg)(5m/s)\\p=25kg.m/s$

For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,

$p=(5kg)(5m/s)\\p=25kg.m/s$

8 0

3 years ago

A tennis player hits a ball at an angle of 50 degrees above horizontal so that it has an acceleration of 12 m/s2. What is the ho

harkovskaia [24]

So we want to know what is the magnitude of the horizontal component of acceleration ah if we know that the overall acceleration a=12 m/s^2 and the angle of overall acceleration and the horizontal acceleration is α=50°. We know that ah=a*cosα. So now it isn't hard to get the horizontal component: ah=12*cos50=12*0.64=7.71 m/s^2. So the correct answer is ah=7.71 m/s^2.

6 0

4 years ago

What equation do we use to solve?

aev [14]

V(y) is the same in all cases so time taken for all cannonballs to go up and down is the same.

4 0

4 years ago