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OlgaM077 [116]
3 years ago
15

Listening to your favorite radio station involves which area of physics?

Physics
1 answer:
lukranit [14]3 years ago
8 0

If I'm not mistaken the answer would be c, vibrations and wave phenomena, because when sound waves hit your ear they go through it to your brain and your mind translates these vibrations to sound. I hope my answer helped you out.

You might be interested in
Try this out. Use a calculator!
Maslowich
Just subsitute and easy
v=55m/s
m=100kg
KE=(0.5)(100kg)(55m/s)^2
KE=(50kg)(3025 m^2/s^2)
KE=151250 J
2nd option
6 0
3 years ago
Which of the Materials are insulators? (select all that apply)
DENIUS [597]
I believe all the insulators would be glass, wood, plastic, and yarn.

But I’m not entirely sure if mechanical pencil lead is an insulator or conductor.

Hope this helps.
3 0
3 years ago
The amplitude of a lightly damped oscillator decreases by 4.2% during each cycle. What percentage of the mechanical energy of th
ra1l [238]

Answer:

V= A ω      maximum KE of object in SHM

V2 / V1 = .958     ratio of amplitudes since ω is constant

KE2 / KE1 = 1/2 m V2^2 / (1/2 m V1^2) = (V2 / V1)^2

KE2 / KE1 = .958^2 = .918

So KE2 = .918 KE1 and .082 = 8.2% of the energy is lost in one cycle

6 0
3 years ago
A large uniform chain is hanging from the ceiling, supporting a block of mass 46 kg. The mass of the chain itself is 19 kg, and
docker41 [41]

Answer:

T = 451.26 N

Explanation:

It is given that,

The mass of block, m = 46 kg

Mass of the chain, m' = 19 kg

Length of the chain, l = 1.9 m

Let T is the the tension in the chain at the point where the chain is supporting the block. It is clearly equal to the product of mass and acceleration.

T=mg

T=46\ kg\times 9.81\ m/s^2

T = 451.26 N

So, the tension in the chain at the point where the chain is supporting the block is 451.26 N. Hence, this is the required solution.

4 0
3 years ago
Four objects are situated along the y axis as follows: a 1.93-kg object is at +2.91 m, a 2.95-kg object is at +2.43 m, a 2.41-kg
Anna [14]

Answer:

0.958 m

Explanation:

So the total mass of the system is

M = 1.93 + 2.95 + 2.41 + 3.99 = 11.28  kg

let y be the distance from the center of mass to the origin. With the reference to the origin then we have the following equation

My = m_1y_1 + m_2y_2 +m_3y_3 + m_4y_4

11.28y = 1.93*2.91 + 2.95*2.43 + 2.41*0 + 3.99*(-0.496) = 10.806

y = \frac{10.806}{11.28} = 0.958 m

So the center of mass is 0.958 m from the origin

3 0
3 years ago
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