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Ratling [72]
3 years ago
15

Describe the action and reaction force pairs involved when an object falls toward earth

Physics
1 answer:
kozerog [31]3 years ago
3 0

Before going to answer this question first we have to understand Newton's third law.

As per newtons third law ,for every action,there is an equal and opposite reaction.The force exerted by a body on other body is called action and force that the body gets in turn is called reaction. They will always act on two different bodies and their direction will be opposite to each other.

As per the question a object is falling under gravity. We are asked to find out the action and reaction forces.

We know that every object having mass will imparts gravitational forces on each other.In this question the earth will apply the force of gravity on the body which is in vertically downward direction.Due to this gravity,the body will fall towards earth with an acceleration of 9.8 m/s^2 .This is the force of action.

The falling object also applies same amount of force on earth in vertically upward direction.But we will not notice any acceleration of earth.It is so because the earth is very massive, so the acceleration produced  by earth is very small.The force exerted by the falling object on earth is the force of reaction.



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A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in
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Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J  

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J  

But  ,

4 * 1875000 = 7500000

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8 0
3 years ago
Read 2 more answers
Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,
adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

               x_cm = 1 /M   ∑ x_{i} m_{i}

               y_cm = 1 /M   ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

    x4=?

   y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

    M = 2.00 + 2.20 + 3.40

    M = 7.6 kg

    0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

     x4 = -6.8 / 6

     x4 = -1,133 m

Axis y

    0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

    y4 = -11/6

    y4 = -1.83 m

body position 4 is (-1,133, -1.83)

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Consider a golf ball with a mass of 45.9 grams traveling at 200 km/hr. If an experiment is designed to measure the position of t
Kipish [7]

Answer:

speed of golf ball is 1.15 × 10^{-30} m/s

and % of uncertainty in speed =  2.07 × 10^{-30} %

Explanation:

given data

mass = 45.9 gram = 0.0459 kg

speed = 200 km/hr = 55.5 m/s

uncertainty position Δx = 1 mm = 10^{-3} m

to find out

speed of the golf ball and  % of speed of the golf ball

solution

we will apply here heisenberg uncertainty principle that is

uncertainty position ×uncertainty momentum ≥ \frac{h}{4\pi }    ......1

Δx × ΔPx  ≥ \frac{h}{4\pi }

here uncertainty momentum ΔPx = mΔVx

and uncertainty velocity = ΔVx

and h = 6.626 × 10^{-34} Js

so put here all these value in equation 1

10^{-3} × 0.0459 × ΔVx =  \frac{6.626*10^{-34}}{4\pi }

ΔVx = 1.15 × 10^{-30} m/s

and

so % of uncertainty in speed = ΔV / m

% of uncertainty in speed =  1.15 × 10^{-30}  / 55.5

% of uncertainty in speed =  2.07 × 10^{-30} %

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