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FromTheMoon [43]
3 years ago
5

Suppose the roller-coaster car in fig.6–41 passes point 1 with a speed of if the average force of friction is equal to 0.23 of i

ts weight, with what speed will it reach point 2? the distance traveled is 45.0 m.
Physics
1 answer:
madreJ [45]3 years ago
8 0

The values in figure 1 shows h1 = 39 m, h2 = 13 m, h3 = 25 with a speed of 1.5m/s

 

The Initial speed u = 1.5m/s.

Vertical distance covered between point 1 and point 2 is (39-13) =26m

The expanse 45 m is equal to a height of (45/6) =7.5 (As frictional force is mg/6)

With the kinematics equation v^2 =u^2 + 2as

V^2 = 1.5^2 + 2*9.8 (26 + 7.5)

V = 25.7 m/s

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Answer:

(A) –14m/s

(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

This involves the knowledge of motion under the action of gravity.

Check below for the full solution to the problem.

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The stars in the sky are organized into groups of stars called constellations which appear near each other in the sky but are no
bezimeni [28]

Answer:

The International Astronomical Union (IAU)  has accepted 88 constellations in the sky.

Explanation:

Constellations has been used since the beginnings of civilizations and each one of them named them as they considered appropiate. It means Greeks' constellations were different than the ones described by Chinese, so it was necessary to gather all these constellations and make a great record with all of them, but there was a problem: Some constellations from different civilizations overlaped because they shared the same stars. There was necessary to put some order on this and that is when in 1922 the International Astronomical Union (IAU) defned a set of 88 moderm constellations  that would become the international standard to look at the night sky. Each one of them is unique and does not share stars with the other constellations.  

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The path the moon's umbra traces across earth's surface is the path of totality.What would you see if you were standing in the p
ivann1987 [24]
You would see the darkest part of the moon
4 0
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It took a student 30 minutes to drive from his home to campus on
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Explanation:

3 0
3 years ago
Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce
jasenka [17]

Answer:

\lambda= 506.25 nm

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

y=\frac{m \lambda D}{a}

Where:

y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1

Solving for λ:

\lambda=\frac{y*a}{mD}

Replacing the data provided by the problem:

\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm

7 0
3 years ago
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