Physics: PLEASE EXPLAIN AND YOU WILL GET BRAINLIST Ms. R is curious if the type of gasoline she uses in her car affects how many miles she can drive on a single tank of gas. (Fuel efficiency). Which of the following would be the BEST scientific question for her to test? 1) How does driving farther affect the type of gas used? 2) How does a Mazda compare to a Civic? 3) How does changing the type of fuel affect the number of miles a car can be driven on a single tank of gas? 4)How does the color of the car

a. The gravitational force of the sun on the earth is Newton.b. The gravitational force of the moon on the earth is Newton.c. The percent of the moon s force to that of the sun s force is 0.57%Given the following data:Mass of sun = kgDistance = meter.Mass of moon = kgDistance = meter.Mass of earth = kgGravitational constant = a. To find the gravitational force of the sun on the earth:Mathematically, Newton s Law of Universal Gravitation is given by the formula:Where:F is the gravitational...

PLEASE EXPLAIN AND YOU WILL GET BRAINLIST Ms.

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joelleygonzalez2778

22.10.2022

Physics

a. The gravitational force of the sun on the earth is $3.54\times 10^{22}$ Newton.

b. The gravitational force of the moon on the earth is $2.0 \times 10^{20}$ Newton.

c. The percent of the moon's force to that of the sun's force is 0.57%

Given the following data:

Mass of sun = $1.99 \times 10^{30}$ kgDistance = $1.55 \times 10^{11}$ meter.Mass of moon = $7.36 \times 10^{22}$ kgDistance = $3.84 \times 10^{8}$ meter.Mass of earth = $6.0\times 10^{24}$ kgGravitational constant = $6.67\times 10^{11}$

a. To find the gravitational force of the sun on the earth:

Mathematically, Newton's Law of Universal Gravitation is given by the formula:

$F = G\frac{M_1M_2}{r^2}$

Where:

F is the gravitational force.g is the gravitational constant.M is the masses of object. r is the distance between centers of the masses.

Substituting the given parameters into the formula, we have;

$F = 6.67\times 10^{-11} \frac{(1.99 \times 10^{30} \times 6.0 \times 10^{24})}{(1.5 \times 10^{11})^2}\\\\F = 6.67\times 10^{-11} \frac{(1.194 \times 10^{55}) }{2.25 \times 10^{22}}\\\\F = \frac{7.96 \times 10^{44} }{2.25 \times 10^{22}}$

F = $3.54\times 10^{22}$ Newton.

b. To find the gravitational force of the moon on the earth:

$F = 6.67\times 10^{-11} \frac{(7.36 \times 10^{22} \times 6.0 \times 10^{24})}{(3.84 \times 10^{8})^2}\\\\F = 6.67\times 10^{-11} \frac{(4.42 \times 10^{47}) }{1.47 \times 10^{17}}\\\\F = \frac{2.95 \times 10^{37} }{1.47 \times 10^{17}}$

F = $2.0 \times 10^{20}$ Newton.

c. To find what percent is the moon's force to that of the sun's force:

$Percent = \frac{2.0 \times 10^{20}}{3.54\times 10^{22}} \times 100$

Percent = 0.57%

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