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Contact [7]
2 years ago
13

1. Describe the three Newton's Law of Motion.

Physics
1 answer:
DIA [1.3K]2 years ago
8 0

Answer:

1st law--Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. The key point here is that if there is no net forceacting on an object (if all the external forces cancel each other out) then the object will maintain a constant velocity. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force.

2nd— The second law explains how the velocity of an object changes when it is subjected to an external force. The law defines a force to be equal to change in momentum(mass times velocity) per change in time. Newton also developed the calculus of mathematics, and the "changes" expressed in the second law are most accurately defined in differential forms. (Calculus can also be used to determine the velocity and location variations experienced by an object subjected to an external force.) For an object with a constant mass m, the second law states that the force F is the product of an object's mass and its acceleration a:

F = m * a

For an external applied force, the change in velocity depends on the mass of the object. A force will cause a change in velocity; and likewise, a change in velocity will generate a force. The equation works both ways.

3rd law-- The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of lift by a wing and the production of thrust by a jet engine.

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Lonnie pitches a baseball of mass 0.20 kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to
suter [353]

Answer:

B) 20N.s is the correct answer

Explanation:

The formula for the impulse is given as:

Impulse = change in momentum

Impulse = mass × change in speed

Impulse = m × ΔV

Given:

initial speed  = 40m/s

Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)

mass = 0.20 kg

Thus, we have

Impulse = 0.20 × (40m/s - (-60)m/s)

Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s

4 0
3 years ago
A book is at rest on a table. Identify the correct free-body diagram for this situation.
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3 years ago
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If a 5 N force pushes a 20 kg mass on a frictionless surface, how fast is the mass going in 5 seconds.
stira [4]

Explanation:

Let me know if you have questions

3 0
3 years ago
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, u=4.52\times 10^5\ m/s

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

a is the acceleration, a=\dfrac{F}{m}

We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

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v=6.45\times 10^5\ m/s

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3 0
3 years ago
Simplifying fractions<br>30 over 54​
Veronika [31]

Answer:

5/9

Explanation:

First, find a common factor for 30 and 54.

30 and 54 can both be divided by 2.

30 divided by 2 = 15

54 divided by 2 = 27

So, your new fraction is 15/27.

However, don't stop there.

15 and 27 can both be divided by 3 as well.

15 divided by 3 = 5

27 divided by 3 = 9

So, your final answer is 5/9.

I hope this helps! :)

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3 years ago
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