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aivan3 [116]
2 years ago
8

1. Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fi

sh really is
Physics
2 answers:
Nutka1998 [239]2 years ago
8 0

Answer:

Closer.

Explanation:

Due to the refraction of light passing through water and air.

pychu [463]2 years ago
3 0

Answer: I believe farther. Correct me if I am wrong

Explanation:

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An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
2 years ago
The orientation of which of the following does not influence the phases of the moon? a. Earth c. Sun b. the moon d. Stars Please
beks73 [17]
I'm pretty sure it's D. The stars don't influence the moon's phases.
6 0
3 years ago
Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is 3.44\times10^{37}\ Kg

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

r=\dfrac{d}{2}

r=\dfrac{15\times9.461\times10^{15}}{2}

r=7.09\times10^{16}\ m

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

M=\dfrac{v^2r}{G}

Put the value into the formula

M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}

M=3.44\times10^{37}\ Kg

Hence, The mass of the massive object at the center of the Milky Way galaxy is 3.44\times10^{37}\ Kg

5 0
3 years ago
Can someone please help me ASAP???!!!
olganol [36]

Answer: The answers are:

Explanation:

1. C

2. A

3. C

4. B

5. D

8 0
1 year ago
A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.
zzz [600]

Answer:

1000 Hz

Explanation:

<em>The frequency would be 1000 Hz.</em>

The frequency, wavelength, and speed of a wave are related by the equation:

<em>v = fλ ..................(1)</em>

where v = speed of the wave, f = frequency of the wave, and λ = wavelength of the wave.

Making f the subject of the formula:

<em>f = v/λ.........................(2)</em>

Also, speed (v) = distance/time.

From the question, distance = 900 m, time = 3.0 s

Hence, v = 900/3.0 = 300 m/s

Substitute v = 300 and λ = 0.3  into equation (2):

f = 300/0.3 = 1000 Hz

6 0
3 years ago
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