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Andrej [43]
3 years ago
10

If, while standing on the bank of a stream, you wished to spear a fish swimming in the water out in front of you, would you aim

above, below, or directly at the observed fish to make a direct hit? If you decided to instead zap the fish with a laser, would you aim above, below, or directly at the observed fish? Defend your answers.
Physics
1 answer:
Serggg [28]3 years ago
3 0

Answer:

<em>a) below the observed position</em>

<em>b) directly at the observed position</em>

<em></em>

Explanation:

If I'm standing on the bank of a stream, and I wish to spear a fish swimming in the water out in front of me, I would aim below the observed fish to make a direct hit. This is because the phenomenon of refraction of light in water causes the light coming from the fish is refract away from the normal as it passes  into the air and into my eyes.

If I'm to zap the fish with a taser, I would aim directly at the observed fish because the laser (a form of concentrated light waves) will refract into the water, taking the same path the light from the fish took to get to my eyes.

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A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.
mash [69]

a) The acceleration of gravity is 4.96 m/s^2

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2

b)

The critical velocity for a satellite orbiting around a planet is given by

v=\sqrt{\frac{GM}{R+h}}

where we have again:

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s

Converting into km/h,

v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

T=\frac{2\pi (R+h)}{v}

where

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

v = 6668 m/s

Substituting,

T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s

d)

One day consists of:

t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

n=\frac{t}{T}=\frac{86400}{8452}=10.2

e)

The escape velocity for an object in the gravitational field of a planet is given by

v=\sqrt{\frac{2GM}{R+h}}

where here we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

Substituting, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s

f)

We can apply again the formula to find the escape velocity for the rocket:

v=\sqrt{\frac{2GM}{R+h}}

Where this time we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=0, because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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