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anyanavicka [17]

3 years ago

14

Cart 1 has a mass of 1 kg. Cart 2 has a mass of 2 kg. They are pushed apart by a spring. The spring exerts 2N of force into cart

1. How much force is exerted into cart 2? Explain.

1 answer:

almond37 [142]3 years ago

5 0

The force exerted on the cart 2 during the collision is 2 N.

The given parameters:

<em>Mass of cart 1 = m1 = 1 kg</em>
<em>Mass of cart 2 = m2 = 2kg</em>
<em>Force applied on cart 1 = 2 N</em>

According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart 1 is equal in magnitude to the force exerted on cart 2 but in opposite direction.

$F_1 = -F_2$

$F_2 = -F_1\\\\F_2 = -2 \ N\\\\|F_2| = 2 \ N$

Thus, the force exerted on the cart 2 during the collision is 2 N.

Learn more about Newton's third law of motion here: brainly.com/question/13874955

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1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant

Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

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First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

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Similarly, the velocity of the stream is, $\vec{v_s}=80\vec{i}$

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

$\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}$

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

$|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph$

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle $\theta$ with the x axis in the third quadrant.

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)$

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

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3 years ago

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Answer:

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Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

$y=\frac{m\lambda D}{a}$ when m is a natural number.

here:

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So we have:

$y_{first}=\frac{633*10^{9}*3.35}{0.0004}$

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b)

Now, if we do m=2 we can find the distance to the second minima.

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Now we need to subtract these distance, to get the width of the first bright fringe :

$y_{second}=10.6-5.3 =5.3 mm$

I hope it heps you!

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