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2 years ago

When two objects move apart from each other, what happens to the gravitational force between them?

Physics

2 answers:

Ierofanga [76]2 years ago

7 0

The answer is C
Because the farther objects are from each other, the weaker the gravitational force is between them.

hichkok12 [17]2 years ago

6 0

C. It decreases

Cuz you know two objects are moving apart from each other the gravitational attraction between them is decreasing. as well as the gravitational force.

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The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

maria [59]

To solve this problem we will start using the concepts related to the electric field, from there we will find the load exerted on the body. Through this load it will be possible to make a sum of forces in balance to find the load that a human supports. Finally with these values it will be possible to find the repulsive force. We will proceed as follows,

The electric field is

$E= \frac{kQ}{R^2}$

Here,

k = Coulomb's Constant

Q = Charge

R = Distance (At this case from the center of mass of the earth to the surface)

Rearranging to find the charge,

$Q = \frac{ER^2}{k}$

Replacing,

$Q = frac{(150)(6.38*10^6)}{8.99*10^9}$

$Q = 6.79*10^5 C$

Since the electric field is directed towards the center of earth, the charge is negative.

PART A) Once the load is found we can proceed to apply the balance of Forces, for which the electrostatic force must be equivalent to the weight, this in order to satisfy the balance, therefore

$F_w = F_e$

$mg = \frac{kQq}{R^2}$

Replacing,

$(62)(9.8) = \frac{(8.99*10^9)(q)(-6.79*10^5)}{(6.38*10^6)^2}$

Solving for q,

$q = -4.056C$

PART B) Finally using the given distance and the values of the found load we can find the repulsive Force, which is

$F =\frac{kq^2}{d^2}$

$F = \frac{(8.99*10^9)(-4.056)^2}{110^2}$

$F = 1.22*10^7N$

PART C) The answer is no. According to the information found, we can conclude that traveling through an electric field is not viable because there is a repulsive force of great magnitude acting on the body.

3 0

2 years ago

3.5 g of a hydrocarbon fuel is burned in a vessel that contains 250. grams of water initially at 25.00 C. After the combustion,

kherson [118]

Answer:

3.) 463 J/g

Explanation:

Heat given by the fuel is used to increase the temperature of the water

so it is given as

$Q = ms\Delta T$

here we will have

m = 250 g

$\Delta T = 26.55 - 25$

now we also know that

$s = 4186$

so we have

$Q = (0.250)(4186)(26.55 - 25)$

$Q = 1622.075 J$

so 3.5 g fuel gives above energy

so per gram of fuel will have total energy given as

$Q = \frac{1622.075}{3.5}$

$Q = 463.45 J/g$

6 0

2 years ago

To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.

Karolina [17]

Answer:

a) T² = ( $\frac{4\pi ^2}{GM}$ ) r³

b) veloicity the dependency is the inverse of the root of the distance

kinetic energy depends on the inverse of the distance

potential energy dependency is the inverse of distance

angular momentum depends directly on the root of the distance

Explanation:

1) for this exercise we will use Newton's second law

F = ma

in this case the acceleration is centripetal

a = v² / r

the linear and angular variable are related

v = w r

we substitute

a = w² r

force is the universal force of attraction

F = $G \frac{m M}{r^2}$

we substitute

$G \frac{m M}{r^2} = m w^2 r$

w² = $\frac{GM}{r^3}$

angular velocity is related to frequency and period

w = 2π f = 2π / T

we substitute

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

the final equation is

T² = () r³

b) the speed of the orbit can be found

v = w r

v = $\sqrt{\frac{GM}{r^3} } \ r$

v = $\sqrt{\frac{GM}{r} }$

in this case the dependency is the inverse of the root of the distance

Kinetic energy

K = ½ M v²

K = ½ M GM / r

K = ½ GM² 1 / r

the kinetic energy depends on the inverse of the distance

Potential energy

U =

U = -G mM / r

dependency is the inverse of distance

Angular momentum

L = r x p

for a circular orbit

L = r p = r Mv

L =

The angular momentum depends directly on the root of the distance

8 0

3 years ago

If i am changing velocity, i must also have _______ and a net _________

Alinara [238K]

if i am changing velocity, i must also have acceleration and a net force

<h2>Newton's first law of motion</h2>

Newton's first law of motion states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

According to Newton's first law of motion, without a force acting on an object, its velocity does not change. The net force acts on an object to change its velocity and cause acceleration.