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777dan777 [17]
3 years ago
14

If, according to Newton’s third law, every reaction has an equal and opposite reaction, then how can motion ever occur? Wouldn’t

every force be canceled out by its reaction force?
Physics
2 answers:
shtirl [24]3 years ago
8 0

No, if you push an object, the other side is also "pulling it".

Motion occurs because of the equal and opposite reaction.

timurjin [86]3 years ago
4 0

Motion could not occur if the action force and the reaction force

both acted on the SAME OBJECT.


But they don't. They act on different objects.


The pusher applies the action force to the thing he's pushing, and

the thing he's pushing applies the reaction force to the pusher.


Newton never said that equal and opposite forces cancel out

when they act on two different objects. Only when they act on

the same object.

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Can anyone plz help me ​
juin [17]

Answer:

D,B,C,A,C

Explanation:

I believe that is the correct answers but it is unclear. I don't think the key for the second last question would let the current flowing so the bulb would be off.

8 0
2 years ago
Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi
kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors \vec{A} and \vec{B} gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors \vec{A} = \vec{P}-\vec{Q} and \vec{B} = \vec{R} - \vec{Q} are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)

\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)

\vec{A} \times  \vec{B} = (A_y B_z - B_y A_z) \  \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}

\vec{A} \times  \vec{B} = ( (-1) * 3 - 1 * (-3) ) \  \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}

\vec{A} \times  \vec{B} = ( - 3 + 3 ) \  \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}

\vec{A} \times  \vec{B} = 0 \  \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}

\vec{A} \times  \vec{B} =(0, -33, 12)

<h2>B)</h2>

We know that \vec{A} and \vec{B} are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

|\vec{A} \times  \vec{B} | = 2 * area_{triangle}

so:

\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}

\sqrt{1233} = 2 * area_{triangle}

35.114= 2 * area_{triangle}

17.55 \ units \  of \ area =  area_{triangle}

5 0
3 years ago
A rail car moving at 10m/s collides with and connect to another stationary car. What is their final velocity?
ki77a [65]

Answer:

read this it might help some

When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater collision force. When a moving object collides with a stationary object of identical mass, the stationary object encounters the greater momentum change.

Explanation:

6 0
3 years ago
If Tom is losing muscle size since he took a break from exercise while he was sick, which exercise principle explains this?
slamgirl [31]
<span>the overload principle hope this helps

</span>
5 0
3 years ago
Alicia is asked to calculate the change in momentum that occurs as she stops her bicycle. This is equivalent to the______ Applie
emmainna [20.7K]
It is the force applied to the bycicle
3 0
3 years ago
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