Dimensional Analysis : 3 days to seconds

Consider Newton's Law of Universal Gravitation: FG= G (m1 m2)/d2 .

bixtya [17]

Answer:

C

Mass is directly proportional to the Force of Gravity. If Mass increases, then the Force of Gravity increases; however, Distance is indirectly (or inversely) proportional to the Force of Gravity. If Distance increases, then the Force of Gravity decreases.

Explanation:

The formula for the force of gravity between two objects is

$F=G\frac{m_1 m_2}{d^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

d is the separation between the two objects

We notice the following:

- F is directly proportional to the masses, $F\propto m_1, m_2$ . This means that if one of the masses increases, then the force between them, F, increases in a proportional way

- F is inversely proportional to the square of the distance, $F\propto \frac{1}{d^2}$ . This means that if the distance between the two objects is increased, the force between them will decrease, and vice-versa.

So, the correct answer is

C

Mass is directly proportional to the Force of Gravity. If Mass increases, then the Force of Gravity increases; however, Distance is indirectly (or inversely) proportional to the Force of Gravity. If Distance increases, then the Force of Gravity decreases.

7 0

3 years ago

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Plot StartRoot 1.5 EndRoot and StartRoot 1.9 EndRoot on the number line to find which inequalities are true. Check all that appl

Vera_Pavlovna [14]

Answer:

A, B, C, E

Explanation:

now gimme thanks please

4 0

2 years ago

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The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

3 years ago

find the mass of a ball on a roof 30 meters high, if the ball's gravitational potential energy is 58.8 joules

Sidana [21]

We are given the gravitational potential energy and the height of the ball and is asked in the problem to determine the mass of the ball. the formula to be followed is PE = mgh where g is the gravitational acceleration equal to 9.81 m/s^2. substituting, 58.8 J = m*9.8 m/s^2 * 30 m; m = 0.2 kg.

3 0

2 years ago

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Nearsightedness can be corrected by using eyeglass lenses that are

Nat2105 [25]

Concaved lenses...........

3 0

2 years ago

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