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dimaraw [331]
3 years ago
10

I NEED HELP LIKE RIGHT NOW!!!!... please

Physics
2 answers:
Maslowich3 years ago
8 0

1.) TRUE

7.) I believe it is FALSE


that's all I got for now

harkovskaia [24]3 years ago
6 0

In my humble opinion . . . . .

1). true

2). a poor question, fishing for "False"

3). true

4). False

5). poor question, fishing for "true"

6). false

7). false

8). false (not totally sure)

9). false

10). true

11). true

12). 600m

13). 0.05 L

14). 400 g

15). kilogram

16). folding

17). solar system

18). matter

19). free-fall

20). smallpox

21). (idk)

22). the continents were once one large land mass

23). the same as her mass on Earth, or anywhere else

24). (idk)

25). magma is one, idk the other

26). a force

27). small

28). formed slowly

29). (idk)

30). distance

31). by collision with a high-speed particle

32). I agree that there is still habitable land remaining on Earth.  But that's not a "therefore" that follows from the first statement.

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A student standing in a canyon yells "echo", and her voice produces a sound wave of frequency of f = 0.61 kHz. The echo takes t
klasskru [66]

Answer:

d=627.9\ m  is the distance from the obstacle of reflection.

wavelength \lamb=0.5279\ m

Explanation:

Given that:

  • frequency of sound, f=610\ Hz
  • time taken for the echo to be heard, t=3.9\ s
  • speed of sound, v=322\ m.s^{-1}

We know,

\rm distance = speed \times time

<em>During an echo the sound travels the same distance back and forth.</em>

2d=v.t

2d=322\times 3.9

d=627.9\ m  is the distance from the obstacle of reflection.

<u>Now the wavelength of sound waves:</u>

\lambda=\frac{v}{f}

\lambda=\frac{322}{610}

\lamb=0.5279\ m

3 0
3 years ago
What is the relationship between the frequency of light and refraction?
Akimi4 [234]

Answer:

The refractive index of a medium is dependent (to some extent) upon the frequency of light passing through, with the highest frequencies having the highest values of n. It also changes the wave speed, frequency, and wavelength.. Although the wave slows down, its frequency remains the same, due to the fact that its wavelength is shorter. When waves travel from one medium to another the frequency never changes. As waves travel into the denser medium, they slow down and wavelength decreases

Explanation:

6 0
3 years ago
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The range of human hearing extends from approximately 20 Hz to 20 000 Hz. Find the wavelengths of these extremes when the speed
soldi70 [24.7K]

Explanation:

Given that,

The range of human hearing extends from approximately 20 Hz to 20 000 Hz. We need to find the wavelengths of these extremes when the speed of sound in air is equal to 338 m/s.

Speed of any wave is given by :

v=f\lambda

\lambda is wavelength of wave

If f = 20 Hz

\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{338}{20}\\\\\lambda=16.9\ m

If f = 20000 Hz

\lambda'=\dfrac{v}{f'}\\\\\lambda'=\dfrac{338}{20000}\\\\\lambda'=0.0169\ m

The smallest wavelength that can be heard is 0.0169 m and the longest wavelength that can be heard is 16.9 m.

7 0
3 years ago
Someone just help me
Zanzabum
C according to my calculations
3 0
3 years ago
Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla
svp [43]

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

So, the average electric field is -30 N/C

6 0
3 years ago
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