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3 years ago

THE NUMERICAL RATIO OF DISPLACEMENT TO DISTANCE IS:

Physics

1 answer:

Tamiku [17]3 years ago

7 0

There's no general rule.

Displacement is the length of a straight line from start to finish, and distance is how far you actually traveled from start to finish.

The only thing we really know is that distance can never be shorter than displacement. So I guess the answer is:

<em>The numerical ratio of displacement to distance is always 1 or less</em>.

(But it has to be written ALL IN CAPITALS.)

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A space vehicle is traveling at 2980 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

Strike441 [17]

Answer:

3054.4 km/h

Explanation:

Using the conservation of momentum

momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor

initial momentum = 14900 M km/h

let v be the new speed of the motor so that the

new momentum = 4Mv and the new momentum of the module = M ( v + 94 km/h )

total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M

initial momentum = final momentum

14900 M km/h = 5 Mv + 93M

14900 km/h = 5v + 93

14900 - 93 = 5v

v = 2961.4 km/h

the speed of the module = 2961.4 + 93 = 3054.4 km/h

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3 years ago

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Sidana [21]

This had to do with gain power and trade inequality business

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3 years ago

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tigry1 [53]

Answer:f 30

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2 years ago

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that: