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MrRa [10]
3 years ago
12

Define Newton’s three laws of motion and how they apply to everyday situations.

Physics
1 answer:
son4ous [18]3 years ago
3 0
<span>1. By Ilkka Cheema<span><span>2. </span>Newton’s 1st Law  The first law of motion sates that an object will not change its speed or direction unless an unbalanced force (a force which is distant from the reference point) affects it. Another name for the first law of motion is the law of inertia. If balanced forces act on an object it doesn’t accelerate or change direction. This means it doesn’t change its velocity and it doesn’t have momentum.</span><span><span>3. </span>Examples of Newton’s 1st Law  If you slide a hockey puck on ice, eventually it will stop, because of friction on the ice. It will also stop if it hits something, like a player’s stick or a goalpost.  If you kicked a ball in space, it would keep going forever, because there is no gravity, friction or air resistance going against it. It will only stop going in one direction if it hits something like a meteorite or reaches the gravity field of another planet.  If you are driving in your car at a very high speed and hit something, like a brick wall or a tree, the car will come to an instant stop, but you will keep moving forward. This is why cars have airbags, to protect you from smashing into the windscreen.</span><span><span>4. </span>Newton’s 2nd Law  The second law of motion states that acceleration is produced when an unbalanced force acts on an object (mass). The more mass the object has the more net force has to be used to move it.</span><span><span>5. </span>Examples of Newton’s 2nd Law  If you use the same force to push a truck and push a car, the car will have more acceleration than the truck, because the car has less mass.  It is easier to push an empty shopping cart than a full one, because the full shopping cart has more mass than the empty one. This means that more force is required to push the full shopping cart.</span><span><span>6. </span>Newton’s 3rd Law The third law of motion sates that for every action there is a an equal and opposite reaction that acts with the same momentum and the opposite velocity.</span><span><span>7. </span>Examples of Newton’s 3rd Law  When you jump off a small rowing boat into water, you will push yourself forward towards the water. The same force you used to push forward will make the boat move backwards.  When air rushes out of a balloon, the opposite reaction is that the balloon flies up.  When you dive off of a diving board, you push down on the springboard. The board springs back and forces you into the air.</span></span>
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4.5 billion km, the average separation between the sun and Neptune (report answer in hours). How long does it take light to trav
Liula [17]

Answer:

t = 4.17 hours

Explanation:

given,

The distance between Sun and Neptune, d = 4.5 billion Km

                                                                         = 4.5 x 10⁹ Km

                                                                          = 4.5 x 10¹¹ m

The velocity of light, c = 3 x 10⁸ m/s

The velocity is always equal to displacement by the time.

                                           <em>V = d / t    m/s</em>

∴                                           t = d / V

                                               = 4.5 x 10¹¹ m / 3 x 10⁸ m/s

                                               = 15,000 s

                                               = 4.17 h

Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h

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What do you know about nuclear energy?​
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A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d
andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x²  - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0
3 years ago
The magnitude of the velocity vector of the car is ∣∣v→∣∣ = 78 ft/s. If the vector v→ forms an angle θ = 0.09 rad with the horiz
kotegsom [21]

Answer:

\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j

Explanation:

The x- and y- components of the velocity vector can be written as following:

\vec{v}_x = ||\vec{v}||\cos(\theta)\^i

\vec{v}_y = ||\vec{v}||\sin(\theta)\^j

Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:

\vec{v} = 78\cos(0.09)\^i + 78\sin(0.09)\^j\\\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j

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2 years ago
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