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3 years ago

If gravity on the earth increased, what affect would it have on the moon

Physics

1 answer:

Rufina [12.5K]3 years ago

6 0

Answer:

If gravity on Earth is increased, this gravitational tugging would have influenced the moon's rotation rate. If it was spinning more than once per orbit, Earth would pull at a slight angle against the moon's direction of rotation, slowing its spin. If the moon was spinning less than once per orbit, Earth would have pulled the other way, speeding its rotation.

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WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car

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3 years ago

Explain why you cannot use mass or volume alone to identify substances.

Alex17521 [72]

Objects can have the same mass (but different compositions). Only mass or volume cannot tell you if the object is solid or volumes) or same volume (but different masses)

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3 years ago

A proton is accelerated to 225V. Its de-Broglie wavelength is:

marin [14]

Answer:

The value is $\lambda = 1.9109 *10^{-12} \ m$

Explanation:

From the question we are told that

The potential of the proton is $V = 225 \ V$

Generally the momentum of the particle is mathematically represented as

$p = \sqrt{ 2 * m * V * e }$

Here e is the charge on the proton with value

$e = 1.60 *10^{-19} \ C$

m is the mass of the proton with value $m = 1.67 *10^{-27} \ kg$

$p = \sqrt{ 2 * (1.67*10^{-27} ) * 225 * 1.6*10^{-19}}$

=> $p = 3.4676 *10^{-22} \ kg \cdot m/ s$

So the de-Broglie wavelength isis mathematically represented as

$\lambda = \frac{h}{p}$

Here h is the Planck's constant with value

$h = 6.626 *10^{-34} \ J\cdot s$

=> $\lambda = \frac{6.626 *10^{-34}}{3.4676 *10^{-22} }$

=> $\lambda = 1.9109 *10^{-12} \ m$

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3 years ago

you are the science officer on a visit to a distant solar system. prior to landing on a planet you measure its diameter to be 1.

Eduardwww [97]

The mass of the star of the distant solar system is 82 * 10²² kg and the mass of the planet of the distant solar system is 8.58 * 10²⁵ kg

The planet stays in its orbit and moves in a centripetal motion due to the gravitational force of attraction between the planet and the star.

Fc = m v² / r

Fg = G M m / r²

Fc = Fg

m v² / r = G M m / r²

v² = G M / r

M = r v² / G

g = Acceleration due to gravity

G = Gravitational constant

M = Mass of star

r = Distance between both bodies

G = 6.67 * 10⁻¹¹ N m² / kg²

r = 8.6 * 10¹¹ m

T = 402 days = 3.47 * 10⁷ s

v = 2 π r / T

v = 2 * 3.14 * 8.6 * 10¹¹ / 3.47 * 10⁷

v = 15.56 * 10⁴ m / s

Ms = 8.6 * 10¹¹ * 15.56 * 10⁴ / ( 6.67 * 10⁻¹¹ )

Ms = 20.1 * 10²⁶ kg

F = G M m / r²

F = m g

Equating,

m g = G M m / r²

Mp = g r² / G

Mp = 63.6 * ( 1.8 * 10⁷ / 2 )² / 6.67 * 10⁻¹¹

Mp = 8.58 * 10²⁵ kg

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11 months ago

Estimate the amount of gasoline that will be consumed in cars in the U.S. in the next thirty years. Assume that the total popula

ioda

Answer:

1182.6 billion gallons of gasoline

Explanation:

For the 350 million population of America, one of two people represents half of this population, which is 135/2 = 67.5 million.

The average distance driven per day by this amount of people is 40 miles, and the average mileage is 25 mpg (miles per gallon).

This means that for the forty miles driven per day, the gallon of fuel used is

==> 25 miles = 1 gallon

40 miles = $x$ gallons.

$x$ = 40/25 = 1.6 gallons, which means that each of the 67.5 million Americans consumes 1.6 gallons of gasoline on an average, per day.

There are 365 days in a week, so in a year, 365 x 1.6 = 584 gallons of fuel is used by one of these 67.5 million Americans.

Which is 584 x 67.5 million = 39.42 billion gallons per year.

In the next 30 years that is 30 x 39.42 billion gallons = 1182.6 billion gallons of gasoline

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3 years ago