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Fynjy0 [20]
3 years ago
9

A roller coaster car of mass 1420 kg slides on a frictionless track starting at a distance 36.83 m above the bottom of a loop 29

m in diameter. The acceleration of gravity is 9.81 m/s 2. If friction is negligible, what is the magnitude of the force of the track on the car when the car is at the top of the loop? Answer in units of N.

Physics
2 answers:
Roman55 [17]3 years ago
3 0

Answer:

6859.079 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (39.83-29)+0^2}\\\Rightarrow v=14.57\ m/s

N=m\frac{v^2}{r}-mg\\\Rightarrow N=1420\left(\frac{14.57^2}{14.5}-9.81\right)\\\Rightarrow N=6859.079\ N

Magnitude of the force of the track on the car when the car is at the top of the loop is 6859.079 N

stiks02 [169]3 years ago
3 0

Answer:

Magnitude of force = 6415.207 N

Explanation:

Given that

m= 1420 kg

h= 36.83 m

d=29 m ,r=14.5 m

g=9.81 m/s²          

From energy conservation

Lets take velocity of particle at top position is V

m g h  - m g d =1/2 m V²    

V² = g(h-d)

V²  = 9.81 x (36.83 - 29)

V=8.76 m/s

At top position

F  + m g = mV² /r

F=\dfrac{mV^2}{r}-mg

By putting the values

F=\dfrac{1420\times 8.76^2}{14.5}-1420\times 9.81\ N

F=−6415.207 N

Magnitude of force = 6415.207 N

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a water bomber flying with a horizontal speed of 85m/s at a height of 3000m drops a load on a fire below. How far in front of th
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Answer:

2081.65 m

Explanation:

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An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act
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Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector  \hat{i}.
The positive y-axis = the northern direction, with unit vector \hat{j}.

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})

The plane's actual velocity is the vector sum of the two velocities. It is
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The magnitude of the actual velocity is
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The angle that the velocity makes north of east is
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Answer:
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