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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

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