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2 years ago

What is the average acceleration given by this graph:

Physics

1 answer:

Zepler [3.9K]2 years ago

5 0

Answer:

Explanation:

There is no graph.

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PLEASE HELP ASAP 2. Two rocks with a mass of 5 kilograms and 10 kilograms, respectively, fall freely from rest near

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i Think momentum

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2 years ago

A student uses the right-hand rule as shown.

Kipish [7]

Answer:

right is the correct answer to the given question .

Explanation:

In this question figure is missing

The main objective right-hand rule to decide the position of the magnetic force on the positive force acting, either the position of the thumb of a right hand with in position of v, the fingers throughout the position of B1, and a right angles throughout the position of F1 to the hand positions.

So $F1 \ =\ q\ v \ B1\ Sin\alpha$

So from the magnetic right hand rule the direction of the magnetic field in front of a wire is right .
All the others options are incorrect because they do not give the direction of the magnetic field in front of a wire is right .

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2 years ago

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Brainliest for whoever is first and right.

devlian [24]

The answer is 9.8, did this last year in AP Science

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2 years ago

The color produced by fireworks is all due to the chemicals inside of the firework itself. Which of these would describe a firew

AlexFokin [52]

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3 years ago

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Water flowing through a cylindrical pipe suddenly comes to a section of the pipe where the diameter decreases to 86% of its prev

masha68 [24]

Answer:

The speed in the smaller section is $43.2\,\frac{m}{s}$

Explanation:

Assuming all the parts of the pipe are at the same height, we can use continuity equation for incompressible fluids:

$\Delta Q=0$ (1)

With Q the flux of water that is $Av$ with A the cross section area and v the velocity, so by (1):

$A_{2}v_{2}-A_{1}v_{1}=0$

subscript 2 is for the smaller section and 1 for the larger section, solving for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}$ (2)

The cross section areas of the pipe are:

$A_{1}=\frac{\pi}{4}d_{1}^{2}$

$A_{2}=\frac{\pi}{4}d_{2}^{2}$

but the problem states that the diameter decreases 86% so $d_{2}=0.86d_{1}$ , using this on (2):

$v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\approx1.35v_{1}$

$v_{2}\approx(1.35)(32)\approx43.2\,\frac{m}{s}$

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3 years ago