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Svetllana [295]
2 years ago
11

You are driving your motorcycle in a circle of radius 75 m on wet pavement. what is the fastest you can go before you lose tract

ion, assuming the coefficient of static friction is 0.20?
Physics
1 answer:
stira [4]2 years ago
3 0

On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s

Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.

Maximum velocity of a road with friction is given by the formula,

v = μRg

where, v is the maximum velocity

μ is the coefficient of static friction

R is the radius of the circle road

g is the acceleration due to gravity

Given,

μ = 0.20

R = 75m

g = 9.8m/s²

On substituting the given values in the above formula,

v = 0.20× 75 ×9.8

v = 147m/s

So, the Maximum velocity of the wet road is 147m/s.

Learn more about Velocity here, brainly.com/question/18084516

#SPJ4

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harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

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Where,

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n_{air} = Refraction index of air

d_w = Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

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d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})

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Once​ Kate's kite reaches a height of 50 ft ​(above her​ hands), it rises no higher but drifts due east in a wind blowing 7 ft d
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Answer:

dz  = 7.136 (answer)

Explanation:

given height  for kate's kite   = 50 ft  (say y)

due to drift it move towards east =  dx = 7 ft  

string maximum length          = 107 ft ( say z)

we have to find change in z  

that is  dz

it will form a right angle triangle for x , y and z  where x is base y is height and z is hypotenuse

so we get according to Pythagoras Theorem

z^2 = x^2+y^2\\...............(i)

by derivative both side  consider y as constant

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from (i)  equation

x = \sqrt{z^2-y^2}

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now put the value and find dz

dz = (\sqrt{120^2-50^2)}7/107\\  =((\sqrt{14400 -2500} )\times7)/107

after solving these  we get

dz  =\frac{109.08\times7}{107}  

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