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katrin [286]
3 years ago
15

Which of the following terms describes the motion of particles during the transmission of wave energy?

Physics
1 answer:
bezimeni [28]3 years ago
7 0

The motion of particles during the transmission of wave energy is periodic. A periodic wave is a continuous oscillating motion that connects with simple harmonic motion. It is usual that periodic wave particles also experiences simple harmonic motion. Additionally, the particles on the surface of the water travel in circular motions forming a wave across the surface.

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¿Cuál es la energía cinética que ejerce un Transmilenio que pesa 19500 kg vacío que se desplaza a 53 Km/h y cual es su energía c
EastWind [94]

Answer:

english?

Explanation:

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3 years ago
If you dropped a ball from a height of 10 m and measured its distance to the ground every 0.1 s, what would the graph of positio
wlad13 [49]

If you dropped a ball from any height, and measured its distance from the ground at any regular interval while it's falling, the graph of that distance versus time would be a graph that curves downward.

-- The ball is falling down.  As time goes on, it gets closer and closer to the ground.  Its remaining distance from the ground keeps decreasing, so the line on the graph slopes down.

-- The speed of the ball keeps increasing (it accelerates) because of the gravitational force on it. As time goes on, it covers more of the remaining distance during each interval than it did in the previous interval. The downward slope of the graph keeps increasing.

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3 years ago
What type of system is earth?<br> A.Closed<br> B.Open<br> C.Connected<br> D.Isolated
adoni [48]
The earth is considered an Open system which is B.
3 0
4 years ago
A lawn mower engine running for 20 m i n does 4, 5 6 0, 0 0 0 J of work. What is the power output of the engine?
VladimirAG [237]

Answer:3800\ W

Explanation:

Given

Lawn mover running for t=20\ min

and does W=4560\times 10^3\ J

We know Power is rate of work i.e.

P=\frac{\text{Work}}{\text{time}}

P=\frac{4560\times 10^3}{20\times 60}

P=3800\ W

Thus Power output is 3800\ W

4 0
3 years ago
A microscope has an eyepiece with a 1.8 cm focal length and a 0.8 cm focal length objective lens. Assuming a relaxed eye, calcul
igomit [66]

Answer:

0.848\ \text{cm}

232.66

Explanation:

N = Near point of eye = 25 cm

f_o = Focal length of objective = 0.8 cm

f_e = Focal length of eyepiece = 1.8 cm

l = Distance between the lenses = 16 cm

Object distance is given by

v_o=l-f_e\\\Rightarrow v_o=16-1.8\\\Rightarrow v_o=14.2\ \text{cm}

u_o = Object distance for objective

From lens equation we have

\dfrac{1}{f_o}=\dfrac{1}{u_o}+\dfrac{1}{v_o}\\\Rightarrow u_o=\dfrac{f_ov_o}{v_o-f_o}\\\Rightarrow u_o=\dfrac{0.8\times 14.2}{14.2-0.8}\\\Rightarrow u_o=0.848\ \text{cm}

The position of the object is 0.848\ \text{cm}.

Magnification of eyepiece is

M_e=\dfrac{N}{f_e}\\\Rightarrow M_e=\dfrac{25}{1.8}\\\Rightarrow M_e=13.89

Magnification of objective is

M_o=\dfrac{v_o}{u_o}\\\Rightarrow M_o=\dfrac{14.2}{0.848}\\\Rightarrow M_o=16.75

Total magnification is given by

m=M_eM_o\\\Rightarrow m=13.89\times 16.75\\\Rightarrow m=232.66

The total magnification is 232.66.

3 0
3 years ago
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