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mixas84 [53]
3 years ago
15

Please help I'm taking a timed test!

Physics
2 answers:
SCORPION-xisa [38]3 years ago
8 0
B. A small rubber ball bouncing near a little girl
xenn [34]3 years ago
8 0

Answer:

B  ......a small rubber ball bouncing near a little girl

Explanation:

All the other answers  arent a type of energy change they aint moving sooo where is the energy  change ...... Nowhere .

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What are tides? the regular daily rises and falls in sea level caused by the gravitational attraction of the Moon and Sun on Ear
klemol [59]

Answer:

the regular daily rises and falls in sea level caused by the gravitational attraction of the Moon and Sun on Earth.

Explanation:

Tides can be defined as the rise and fall of water level in water bodies such as lakes and oceans due to the gravitational force of attraction exerted by the moon on earth. The side closest to the moon creates a bulge of water known as high tide. Low tides are generally experienced when a sea level is not within the bulge.

Generally, the gravitational pull of the Moon cause visible changes on planet Earth's surface.

This ultimately implies that, the pull of the Moon's gravity causes high and low tides on planet Earth's surface.

The various types of ocean tides based on the position of the Earth, Moon and the Sun are;

I. Neap tides.

II. Spring tides.

III. Low tide.

IV. High tide.

V. Brown tide.

VI. Rip tide.

VII. Red tide.

6 0
3 years ago
What is the frequency of radiation whose wavelength is 11.5 a0 ?
irakobra [83]

Answer:

The frequency of radiation is 2.61 \times 10^{17} s^{-1}

Explanation:

Given:

Wavelength \lambda = 11.5 \times 10^{-10} m

Speed of light c = 3 \times 10^{8} \frac{m}{s}

For finding the frequency of radiation,

  c = f \lambda

  f = \frac{c}{\lambda}

  f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }

  f = 2.61 \times 10^{17} s^{-1}

Therefore, the frequency of radiation is 2.61 \times 10^{17} s^{-1}

4 0
3 years ago
What is the displacement from 4 to 10 seconds? How do you know?
Strike441 [17]
For velocity vs time graphs, you can calculate displacement by finding the area between the line and the x-axis (time line). You should be able to recognize that space between the lines as a triangle, trapezoid, square, or rectangle and use the respective area formulas for these shapes.

See attached photo for worked out solution.

4 0
3 years ago
Scientists are investigating how well different microphones capture and record sounds. They use tools that show how loud the hig
inna [77]

Answer:

Which microphone captures a singer's low sounds the loudest?

Explanation:

Scientific questions are the basis of any scientific experiment. An observation is followed by a scientific question. In this question, scientists are investigating how well different microphones capture and record sounds by using tools that show how loud the high and low sounds are and a tool that shows how much noise is recorded along with the actual sounds.

Hence, a scientific question should be that which will cover the whole aspect of the investigation i.e. the question should contain a general scope of the problem. Hence, a scientific question that could be included in this investigation is:

Which microphone captures a singer's low sounds the loudest?

This is because this question contains every part being investigated i.e. How different microphones capture sounds and records them (whether low or high).

7 0
3 years ago
A loudspeaker having a diaphragm that vibrates at 910 Hz is traveling at 75.0 m/s directly toward a pair of holes in a very larg
Alina [70]

To solve this problem we need to apply doppler equation,

Our values are:

f_0 = 910Hz\\c = 75m\\v=344m/s\\\theta= 12.9\°\\

Doppler equation is given by,

f= \frac{v}{v-c} f_0

Substituting,

f= \frac{344}{344-75}*910

f = 1163Hz

\lambda = \frac{344}{1163}

Wavelenght is equal to,

\lambda = \frac{c}{f}

\lambda = 0.2956

We can now find the distance by,

d= \frac{1}{2}\frac{\lambda}{sin\theta}

d=\frac{1}{2}\frac{0.2956}{sin(12.9)}

d= 0.66m

b) We can find at what angles would sound first cancel through the same equation of wavelenght,

\lambda = \lambda{c}{f} = \frac{344}{1163}

d* sin\theta = \frac{1}{2}*\lambda

\theta = arcsin( 0.5*(344/1163)/0.66)=12.94\°

5 0
3 years ago
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