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Rufina [12.5K]
2 years ago
5

Object A and Object B are 100 meters apart. If Object A gains some

Physics
1 answer:
satela [25.4K]2 years ago
7 0

The gravitational force between the two objects A) It increases.

Explanation:

The gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2} (1)

where

G is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between the objects

In this problem, object A and object B are initially at a distance of

r = 100 m

And at that distance, the force between them is

F

Later, object A gains some mass. We notice from eq.(1) that the gravitational force is directly proportional to the mass: therefore, if the mass of either of the two objects increases, then the gravitational force between them also increases. Therefore, the new force will be larger than the original force:

F' > F

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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Density is a measure of an object's _____ per unit _____.
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<span>Density is a measure of an object's </span>mass per unit of volume

Which means that it shows how much mass is contained within a volume of something.
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3 years ago
Which of the following processes charges an object by friction
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An example is when u rub your pen on your hair hard that is friction
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3 years ago
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An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5
Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

5 0
3 years ago
A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back
elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=9.8 m/s^2

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates=\mu mg=0.202\times 9.8\times 35

Force by which box accelerates=62.286 N

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

ma=\mu mg

a=\mu g=9.8\times 0.202=1.9796 m/s^2

Hence, the truck can have maximum acceleration before the box slide=1.9796 m/s^2

3 0
3 years ago
A bicyclist travels 4.5 km west, then travels 6.7 km at an angle 27.0 degrees South of West. What is the magnitude of the bicycl
Dimas [21]

Answer:

<em>10.90km</em>

Explanation:

Magnitude of the total displacement is expressed using the equation

d = √dx²+dy²

dx is the horizontal component of the displacement

dy is the vertical component of the displacement

dy = -6.7sin27°

dy = -6.7(0.4539)

dy = -3.042

For the  horizontal component of the displacement

dx = -4.5 - 6.7cos27

dx = -4.5 -5.9697

dx = -10.4697

Get the magnitude of the bicyclist's total displacement

Recall that: d = √dx²+dy²

d = √(-3.042)²+(-10.4697)²

d = √9.2538+109.6146

d = √118.8684

<em>d = 10.90km</em>

<em>Hence the magnitude of the bicyclist's total displacement is 10.90km</em>

<em></em>

6 0
2 years ago
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