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3 years ago

In a surface wave (such as a water wave), particles __________.

Physics

2 answers:

soldi70 [24.7K]3 years ago

8 0

Answer

In a surface wave (such as a water wave), particles <u>like water molecule undergoes circular motion.</u>

Explanation:

surface waves are the waves which does not possess transverse or longitudinal nature. In surface waves (water waves) all the particles of the bulk medium(water) does not moves in perpendicular direction or in parallel direction with respect to energy transport direction. In surface wave only surface particles moves in circular motion.

For example when we through a stone in water that time only surface particles moves in circular direction and it creates ripples.

weqwewe [10]3 years ago

7 0

Particles similar to a water wave will travel away from the source. On the Earth's surface this would be shown as an aftershock. On the surface of water, the waves would travel in ripples.

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3 years ago

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How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

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11 months ago

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

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3 years ago

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Goshia [24]

Answer:

The output out be 200

Explanation:

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ra1l [238]

Answer:

The distance between the camera and the rock is 836.6 cm

Explanation:

A right triangle is formed where the hypotenuse (h) is the distance between the rock and the camera. One of the leg (l) is the distance between the camera and the surface. The angle between the hypotenuse and this leg is α = 90° - 13.69° = 76.31°. By definition:

cos α = adjacent/hypotenuse

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3 years ago