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3 years ago

Gravitational force is reduced bybetween objects.

1 answer:

6 0

Answer: When the distance between the two object triples then the gravitational force is reduced by one ninth (1/9)th between objects. Gravitational force is inversely proportional to the square of the distance of two object.

Explanation: Therefore, the gravitational force is reduced by the square of the distance between objects.

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Consider this situation: Four ropes, each attached to the end

faust18 [17]

The forces acting on the elevator are:

Gravity force

Tension force

Air resistance

Explanation:

Let's go through each of the forces listed and see which ones are acting on the elevator.

Normal force: NO. The normal force is a force exerted by a surface whenever there is another object "pushing" on it. For instance, when a box is at rest on a table, the box is "pushing" on the table (due to its weight), and the table "pushes back" on the box, upward, in order to balance its weight: this is the normal force. In this case, the elevator is lifted, so it is not pushing on anything, therefore there is no normal force.
Gravity force: YES. The force of gravity acts on every object located in the gravitational field of the Earth; it pulls downward, and its magnitude is $mg$ , where m is the mass of the object and g is the acceleration of gravity.
Applied force: NO. Here there is no applied force, since there is nobody "pushing" or "pulling" the elevator.
Friction force: NO. As we are considering the forces on the elevator, and the elevator is not sliding against any surfaces, there is no force of friction. (The force of friction acts whenever there are two surfaces sliding against each other, which is not the case here)
Tension force: YES. The tension force is the force exerted by a rope or a string when pulling an object. In this case, there are four ropes pulling the elevator, therefore there are 4 forces of tension acting on the elevator, upward.
Air resistance: YES. As the elevator is moving through the air, the interaction between the molecules of air with the surface of the elevator produces a force (called air resistance) that "resists" the motion of the elevator, therefore pushing downward. However, the magnitude of this force is negligible in this case.

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

5 0

3 years ago

a specimen of oil having an initial volume of 5000cm³ is subjected to a pressure of 10⁴N/m² and the volume decreases by 0.20cm³.

inessss [21]

Answer:

B = 2.5 10⁸ Pa

Explanation:

The volume modulus is defined by

B = $- \frac{P}{ \frac{\Delta V}{V} }$

The negative fate is for the module to be positive since the volume change is negative

It is not necessary to reduce the volumes to the SI system, since they are both in the same units

B = $- \frac{10^4}{ \frac{-0.20}{5000} }$ = $\frac{10^4}{4 \ 10^{-5} }$

B = 2.5 10⁸ Pa

4 0

3 years ago

Lightwaves travel from the air into a lens made of glass. Their velocity decreases as they enter the glass. How does this affect

Papessa [141]

The waves become longer but slower

4 0

3 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

An astronaut having mass 320 kg with equipment included is attempting an untethered space walk. The astronaut is initially at re

ExtremeBDS [4]

This can be solved using momentum balance, since momentum is conserved, the momentum at point 1 is equal to the momentum of point 2. momentum = mass x velocity
m1v1 = m2v2
(0.03kg x 900 m/s ) = 320(v2)
v2 = 27 / 320
v2 = 0.084 m/s is the speed of the astronaut

7 0

3 years ago

Gravitational force is reduced bybetween objects.​

Gravitational force is reduced bybetween objects.