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goldfiish [28.3K]
2 years ago
6

Newton’s laws of motion are an inherent part of our everyday life. Starting from the beginning of the day till the end, Newton’s

laws find their applications in a number of activities we do.
Physics
1 answer:
12345 [234]2 years ago
6 0

1st

Newton's laws of motion are three basic laws of classical mechanics that describe the relationship between the motion of an object and the forces acting on it. These laws can be paraphrased as follows: Law 1. A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force.

2nd

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

3rd

His third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.

hope this helps u :)

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A playground merry-go-round of radius r = 1 m has a moment of inertia of i = 240 kg*m^2. and is rotating at a rate of ω = 8 rev/
Ghella [55]
When the child jumps onto the merry-go-around the moment of inertia of the system changes. If we consider the child to be point-like mass then its moment of inertia would be:
I_{ch}=mr^2
We get the new moment of inertia by simply adding the child's moment of inertia to the old moment of inertia.
I_{new}=I_{old}+I_{ch}=240+35(1)^2=275 $kgm^2
Since there is no force mention we must assume that angular momentum is conserved.
L=const.\\ L=I_{old}\omega_0=I_{new}\omega'\\ \omega'=\frac{I_{old}\omega_0}{I_{new}}
When we plug in all the numbers we get:
\omega'=\frac{240\cdot8}{275}=6.98 \ \frac{rev}{min}

3 0
3 years ago
To determine the amplitude of a wave, you would need to know the . A) distance from the crest to the trough. B) maximum displace
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3 years ago
Read 2 more answers
The electric potential at a position located a distance of 20.7 mm from a positive point charge of 8.60×10-9C and 15.1 mm from a
max2010maxim [7]

Answer:

q2 = -4.35*10^-9C

Explanation:

In order to find the values of the second charge, you use the following formula:

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}       (1)

V: electric potential = 1.14 kV = 1.14*10^3 kV

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1: charge 1 = 8.60*10^-9 C

q2: charge 2 = ?

r1: distance to the first charge = 20.7mm = 20.7*10^-3 m

r2: distance to the second charge = 15.1mm

You solve the equation (1) for q2, and replace the values of the other parameters:

q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C

The values of the second charge is -4.35*10^-9C

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Using an example, explain how small atoms are.
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Korvikt [17]

Answer:

Heat or thermal energy!!!

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