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HACTEHA [7]
3 years ago
12

The roller-coaster car shown in fig. 6-41 (h1 = 45 m, h2 = 16 m, h3 = 26 m), is dragged up to point 1 where it is released from

rest. assuming no friction, calculate the speed at points 2, 3, and 4.
Physics
1 answer:
kirill [66]3 years ago
7 0

There are many ways to solve this but I prefer to use the energy method. Calculate the potential energy using the point then from Potential Energy convert to Kinetic Energy at each points.

PE = KE

From the given points (h1 = 45, h2 = 16, h<span>3  </span>= 26)

Let’s use the formula: 

v2= sqrt[2*Gravity*h1]  where the gravity is equal to 9.81m/s2

v3= sqrt[2*Gravity*(h1 - h3 )] where the gravity is equal to 9.81m/s2

v4= sqrt[2*Gravity*(h1 – h2)] where the gravity is equal to 9.81m/s2

Solve for v2

v2= sqrt[2*Gravity*h1]      

    = √2*9.81m/s2*45m

v2= 29.71m/s

v3= sqrt[2*Gravity*(h1 - h3 )   

    =√2*9.81m/s2*(45-26)

    =√2*9.81m/s2*19 

v3=19.31m/s

v4= sqrt[2*Gravity*(h1 – h2)]        

    =√2*9.81m/s2*(45-16)

    =√2*9.81m/s2*(29)

v4=23.85m/s

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Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

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Answer:

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Explanation:

It is given that,

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Energy of a gas molecules is given by :

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T=\dfrac{2\times P\times t}{3Nk}

T=\dfrac{2\times 125\times 360}{3\times 6\times 10^{26}\times 1.38\times 10^{-23}}

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Hello!

\large\boxed{F = 100N}

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F = 5 · 20

F = 100N

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