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9966 [12]
3 years ago
12

Which best describes what happens when light traveling through air enters water at an angle?

Physics
2 answers:
vovangra [49]3 years ago
3 0
I think it is "A'' 'It moves along straight lines in air and changes direction when it enters water.
galben [10]3 years ago
3 0

The correct choice is

A. It moves along straight lines in air and changes direction when it enters water.

This is because of phenomenon of refraction which  a ray of light experience when it moves from one medium to another medium. The light ray bends away  from the normal to the surface if it travels from denser to rarer medium. The light ray bends towards from the normal to the surface if it travels from rarer to denser medium.

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NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 29.5 m/s. Assume
insens350 [35]

Answer:

boy if you dont get t f outa here

4 0
3 years ago
4. A high school athlete runs 400 m in 52.20 sec. What is his speed in mph?<br> 14100<br> 1110
Serggg [28]

Answer:

28,800m/p second

Explanation:

Calculate the distance per second so, 400m/50 s= 8m/p second now knowing the speed/hour and knowing an hour has 3,600 seconds,multiply it by 8 then you will get 28,800m/p second, or 28.8km/h

3 0
3 years ago
What could contribute to an inaccurate vital sign reading?
amid [387]
<span>Lack of training in getting the vital sign or worst, not knowing the right way to take the vital sign could contribute to an inaccurate vital sign reading. For example, if you are tasked to get the respiration of the patient, the rule is to count inhale and exhale as one. But if you were not able to know this rule, and you counted inhale as one and exhale as another, this could impair the vital reading. </span>
7 0
3 years ago
En la Tierra un volcán puede expulsar rocas verticalmente hasta una altura máxima H. A) ¿A qué altura (en términos de H) llegarí
Nonamiya [84]

A) 2.64 H

The maximum height that the expelled rock can reach can be found by using the equation:

v^2-u^2 = 2gd

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g is the acceleration of gravity

d is the maximum height

Solving for d,

d=\frac{-u^2}{2g}

We see that the maximum heigth is inversely proportional to g. On the Earth,

d=H and g=g_e = -9.81 m/s^2

So we can write:

\frac{H}{H'}=\frac{g_m}{g_e}

where H' is the maximum height reached on Mars, and g_m = -3.71 m/s^2 is the acceleration of gravity on Mars. Solving for H',

H' = \frac{g_e}{g_m}H = \frac{-9.81}{3.71}H=2.64 H

B) 2.64T

The time after which the rock reaches the maximum height can be found by using

v=u+gt

where

v = 0 is the velocity at the maximum height

u is the initial velocity

Solving for t,

t=\frac{v-u}{g}

The total time of the motion is twice this value, so:

t=2\frac{v-u}{g}

So we see that it is inversely proportional to g.

On the Earth, t = T. So we can write:

\frac{T}{T'}=\frac{g_m}{g_E}

where T' is the total time of the motion on Mars. Solving for T',

T' = \frac{g_e}{g_m}T=\frac{-9.81}{-3.71}T=2.64T

4 0
3 years ago
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