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miv72 [106K]
3 years ago
14

Plss help me fast and finish all directions. Correct Drawing/picture plss. Make sure the drawing is clear so I can snip it and p

aste it.
Draw a Roller Coaster Track that starts with one large hill, then is followed

by a Valley and another, smaller hill. Draw a cart in four positions on the track as outlined below.

i. Draw the first cart at the top of the first hill. Label it A.

ii. Draw the second cart going down the first hill into the valley. Label it B.

iii. Draw the third cart at the bottom of the valley. Assume that the height of the cart in this position is zero. Label it C.

iv. Draw the last cart at the top of the second, smaller hill. Label it D.
Physics
1 answer:
jeka943 years ago
5 0

Answer:

Ive left an image here for use, I hope its helpful

Explanation:

I have left two images and i hope i am answering your question.

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What is the primary atomic characteristic that Mendeleev used to organize the periodic table?
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What is the relationship among amplitude,crest,and trough?
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A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang
Dafna11 [192]

Answer:

19.5°

Explanation:

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1) E=\frac{1}{2}mv^2+mgh

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) E=\frac{1}{2}mv^2

The energy when the mass comes to a stop will be:

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Setting equations 2 and 3 equal and solving for height h will give:

4) h=\frac{v^2}{2g}

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) cos\phi=\frac{l-h}{l}

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) \phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})

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A long, straight wire carrying a current of 2.60 A is placed along the axis of a cylinder of radius 0.500 m and a length of 4.25
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<h2>Answer:</h2>

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<h2>Explanation:</h2>

Since the current carrying wire is placed along the axis of the cylinder, according to the right hand rule, the magnetic field will be tangent to the surface of the cylinder. Therefore, there is no magnetic field through the cylinder.

Remember that the magnetic flux through a given area is the total magnetic field passing through that area. Since there is not magnetic field through the cylinder, the total magnetic flux is therefore zero (0).

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