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Andrew [12]
3 years ago
8

(Use the Pythagorean theorem to answer the question.) An airplane takes off going straight west at 340 km/h for 1 hour, then tur

ns and heads south for another hour at 360 km/h before reaching its final destination. What is the airplane's displacement?
Physics
1 answer:
Ksju [112]3 years ago
4 0
South is perpendicular to West, so the plane's route forms a right triangle, and you can use Good Old Pythagoras to calculate the length of the hypotenuse.

The length of the displacement is   √(the west piece² + the south piece²)   .

That's      √ [ (340km)² + (360km)² ]

           =  √ [ (115,600) km²  +  (129,600) km² ]

           =   √ 245,200 km²

           =     495 km 

To be technical, Displacement is a vector, so we would need to
calculate its direction too.

Naturally, the plane winds up roughly southwest of where it took off.

You'd want to find the angle whose tangent is  (360/340) = about 1.059 .
The direction of the Displacement is that angle south of west. (about 46.6 degrees)
        
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Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
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(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

U=-\frac{GMm}{R+h}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

U_A=-\frac{GMm}{R+h_A}

while potential energy of satellite B is

U_B=-\frac{GMm}{R+h_B}

Therefore the ratio of the potential energy of satellite B to that of satellite A is

\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

K=\frac{GMm}{2(R+h)}

So the kinetic energy of satellite A is

K_A=\frac{GMm}{2(R+h_A)}

while kinetic energy of satellite B is

K_B=\frac{GMm}{2(R+h_B)}

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}

which is identical to before, so it  gives

\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473

(c) Satellite B

The total energy of a satellite in orbit is given by

E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) 1.03\cdot 10^7 J

We have to calculate the total energy of each satellite.

Given:

G=6.67\cdot 10^{-11}

M=5.98\cdot 10^{24} kg

m = 12.0 kg

R+h_A = 6370 km+5920 km=12290 km = 12.3 \cdot 10^6 m

R+h_B = 6370 km+19600 km=25970 km = 26.0 \cdot 10^6 m

We find:

E_A = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(12.3\cdot 10^6)}=-1.95\cdot 10^{7} J

E_B = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(26.0\cdot 10^6)}=-9.2\cdot 10^{6} J

So the difference in total energy is

E_B-E_A = -9.2\cdot 10^6 - (-1.95\cdot 10^7) =1.03\cdot 10^7 J

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