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2 years ago

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An airplane, flying horizontally at an altitude of 3 miles, passes directly over an observer. If the constant speed of the air

plane is 240 miles per hour, how fast is its distance from the observer increasing 45 seconds later? Hint: Note that in 45 seconds (three fourths times StartFraction 1 Over 60 EndFraction equals StartFraction 1 Over 80 EndFraction hour ), the airplane goes 3 miles.

1 answer:

natima [27]2 years ago

6 0

I’m not sure if this helped you so here you go

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What is night vision pls tell fast

mario62 [17]

Answer:

Night vision is when you can see in the dark or at night like owls

Explanation:

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2 years ago

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A wagon is accelerating down a hill. Which statement is true?

Ira Lisetskai [31]

Potential energy decreases and kinetic energy increases.

Potential energy is related to the height, since the wagon is going downhill, height decreases and potential energy decreases.

Kinetic energy is related to the speed, since the wagon is speeding up, kinetic energy increases.

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3 years ago

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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3 years ago

Heeeeellllpppppp meeee please

solniwko [45]

Answer:

i cant solve this!

Mybe i can solve another question!

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2 years ago

If two children, with masses of 16 kg and 25 kg, sit in seats opposite one another in a rotating merry-go-round with an arm leng

Jlenok [28]

Answer:

$92.25 kgm^2$

Explanation:

Assume both children bodies are point particles. The total moment of inertia about the rotation axis of 2 points particles of mass 16 kg and 25 kg at 1.5 m arm length is

$\sum I = m_1r_1^2 + m_2r_2^2$

where $m_1 = 16 kg, m_2 = 25 kg$ are the masses of 2 children $r_1 = R-2 = 1.5m$ are their distance to the center of rotation

$\sum I = 16*1.5^2 + 25*1.5^2 = 1.5^2(16 + 25) = 92.25 kgm^2$

4 0

3 years ago