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4 years ago

14

A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically dr

ops that distance before the track curves forward. If friction is neglected, what would be the speed of the roller coaster at the bottom of the drop?
17.60 m/s
24.90 m/s
49.79 m/s
70.42 m/s

1 answer:

yan [13]4 years ago

7 0

Answer:

49.79 m/s

Explanation:

Given:

Initial velocity of the roller coaster is, $u=0\ m/s$

Vertical drop or the displacement of the roller coaster is, $y=-126.5\ m$

The displacement is negative as the motion is in downward direction.

Now, as the motion is in vertical direction only, the acceleration of the roller coaster will be due to gravity acting in the downward direction.

So, the acceleration of the roller coaster is, $a=g=-9.8\ m/s^2$

Now, using the following equation of motion:

$v^2=u^2+2ay$

Where, 'v' is the velocity of the roller coaster at the bottom.

Plug in all the given values and solve for 'v'. This gives,

$v^2=0^2+2(-9.8)(-126.5)\\\\v^2=2479.4\\\\v=\sqrt{2479.4}\\\\v=49.79\ m/s$

Therefore, the speed of the roller coaster at the bottom of the drop is 49.79 m/s.

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Answer:

a). M = 20.392 kg

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Explanation:

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If the block is ready to slide,

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2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

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M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$ .............(iii)

Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

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We got, N = mg cos θ

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$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$ (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$ .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

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$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

$a_M= 0.28 \ m/s^2$

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Answer:

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4 atoms of O: 4x16=64 g/mol

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