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olga_2 [115]
3 years ago
13

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2

. The car makes it one quarter of the way around the circle before it skids off the track. From these data, determine the coefficient of static friction between the car and track.
Physics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

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Sergio [31]
Hi I am the guy to get mad at
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3 years ago
Solve for work when
BlackZzzverrR [31]

So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

  • If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
  • If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.
<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

\boxed{\sf{\bold{W = F \times s}}}

If the Angle Effect on Work

\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}

With the following condition:

  • W = work that done by object (J)
  • F = force that applied (N)
  • s = shift or distance (m)
  • \sf{\theta} = angle of elevation (°)

<h3>Solution</h3>

We know that :

  • F = force that applied = \sf{1.41 \times 10^4} N
  • s = shift or distance = 84.9 m
  • \sf{\theta} = angle of elevation = 45°

What was asked ?

  • W = work that done by object = ... J

Step by step :

\sf{W = F \times s \times \cos(\theta)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}

\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}

\sf{W = 59.8545 \sqrt{2}}

\boxed{\sf{W \approx 84.65 \: J}}

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

3 0
1 year ago
Which phenomenon would cease to exist in the absence of this axial tilt?
konstantin123 [22]
If the Earth didn't tilt then we wouldn't have seasons.
8 0
3 years ago
A sphere of metal has a mass of 1431 g and a diameter of 5.80 cm. What is the density of metal in g/cm3
Dmitriy789 [7]

Answer:

see below

Explanation:

You will need t find the volume of the sphere

 4/3 pi r^3    divide into the mass

1431 / (4/3 pi (5.8)^3) = 14 gm /cm^3

4 0
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If a 1.0-kg-mass block is on the left cap, how much total mass must be placed on the right cap so that the caps equilibrate at e
Dennis_Churaev [7]

<span>The total mass that should be placed in the right cap so that the caps equilibrate at equal height is also 1 kg. if equilibrium should be maintained the force in each side should cancel out, so to balance a 1kg mass, a 1 kg mass should also be place on the opposite direction</span>

3 0
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