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MA_775_DIABLO [31]
1 year ago
11

Two objects move toward each other because of gravitational attraction. As the objects get closer and closer, the force between

them _____. A. remains constant B. decreases C. increases
Physics
2 answers:
bearhunter [10]1 year ago
5 0
<span>Two objects move toward each other because of gravitational attraction. As the objects get closer and closer, the force between them increases. </span>
never [62]1 year ago
5 0

Answer:

C. increases

Explanation:

Law of Universal Gravitation

This law establishes that bodies, by simply having mass, experience a force of attraction to other bodies with mass, called gravitational force or gravitational force.

The gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them.

F_{g} =G(\frac{M*m}{r^{2} } ) Formula (1)

where:

G is the universal gravitation constant, G = 6.67 · 10-11 N · m² / kg²

M and m are the masses of the bodies that interact  (kg)

r is the distance that separates them. (m)

Problem development

As objects get closer and closer the distance (r) that separates them decreases and because M, m and G remain constant, in formula (1):

F_{g} =G(\frac{M*m}{r^{2} } )

, if r decreases then Fg increases.

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Plug variables expressed in SI units in the kinematic equation given in article: a = -v0^2/(2sg). What value of g you get as exp
Elanso [62]

Answer:

1) acceleration is increased by a factor of four 4X

2) the acceleration increases a factor of 2X

3) the correct answer of 400g

Explanation:

This is a kinematics exercise, where you use the velocity equation to obtain the acceleration, with the final velocity equal to zero.

           v² = v₀² + 2 a x

           0 = v₀² + 2 a x

           a = - v₀² / 2 x

           

In the case of wanting to give the acceleration as a function of g, we can find the relationship between the two quantities

         a / g = - v₀² / (2 x g)

Let's answer the different questions about this equation

1. The initial velocity is doubled, how much the acceleration is worth

           

       a/g = - (2v₀) 2 / 2xg

       a = 4 (-v₀² / 2xg) g

acceleration is increased by a factor of four 4X

2. if the stopping distance is reduced by 2, that is, x = x₀ / 2

we substitute

        a/g = (- v₀² / 2g) 2/x

         

        a =2  (-v₀² / 2x₀g)  g

       

therefore the acceleration increases a factor of 2X

3. the initial velocity of the hockey player is v₀ = 20 m / s and the stopping distance is

x = 5cm = 0.05m

we calculate the acceleration

        a / g = - 20² / (2 0.05)

        a / g = - 4000 / g

        a / g = - 4000 / 9.8 = 408

        a = 408 g

the correct answer of 400g, the value matches exactly if g = 10 m / s2 is taken

6 0
2 years ago
Susie walks 3 blocks north to the local CVS store, then 4 blocks east to her grandmother’s house. She then walks 2 blocks west a
Slav-nsk [51]

Answer:

Suzie is 3 blocks north of where she started

Explanation:

Displacement is the minimum distance between the initial and final point of motion.

Here, Suzie first walks 3 blocks north. From there she walks 4 blocks east. Then 2 blocks to the east then 2 blocks north and then 2 blocks east. She covered 4 blocks east toward west. This is the same distance she covered traveling east. But she is 2 blocks north. From there she traveled a block south to the pizzeria and another block to her friends house. She covered the two block she had traveled north.

Hence, Suzie is 3 blocks north of where she started.

7 0
1 year ago
An airplane travels horizontally at a constant velocity v. An object is dropped from the plane and one second later another obje
Delvig [45]

Answer:

the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

Explanation:

Since airplane is moving horizontally with constant speed v

so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane

so we can say that two object when dropped after some interval of time then they always lie in same vertical line

now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other

So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

8 0
2 years ago
A 600g toy train completes 10 laps of its circular track in 1 min 20s. If the radius of the track is 1.2 m, Find the centripetal
Lynna [10]

Wow !  This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.

The centripetal acceleration of any object moving in a circle is

                          (speed-squared)  /  (radius of the circle)  .

Notice that we won't need to use the mass of the train.

We know the radius of the track.  We don't know the trains speed yet,
but we do have enough information to figure it out.  That's what we
need to do first.

Speed  =  (distance traveled) / (time to travel the distance).

Distance = 10 laps of the track.   Well how far is that ? ? ?

1 lap = circumference of the track = (2π) x (radius) =  2.4π  meters

10 laps =  24π  meters.

Time = 1 minute 20 seconds  =  80 seconds

The trains speed is  (distance) / (time)

                               =  (24π meters) / (80 seconds)

                               =        0.3 π  meters/second .

NOW ... finally, we're ready to find the centripetal acceleration.

                                 <span> (speed)²  /  (radius)

                           =    (0.3π m/s)²  /  (1.2 meters)

                           =    (0.09π m²/s²)  /  (1.2 meters)

                           =    (0.09π  /  1.2)   m/s²

                           =          0.236  m/s²  .        (rounded)

If there's another part of the problem that wants you to find
the centripetal FORCE ...

Well,       Force = (mass) · (acceleration) .

We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force.       </span>
4 0
1 year ago
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly abov
Rudik [331]

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

5 0
1 year ago
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