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Nonamiya [84]
3 years ago
5

A highway curves to the left with radius of curvature of 36 m and is banked at 28 ◦ so that cars can take this curve at higher s

peeds. Consider a car of mass 1717 kg whose tires have a static friction coefficient 0.87 against the pavement. top view R = 36 m 28 ◦ rear view µ = 0.87 How fast can the car take this curve without skidding to the outside of the curve? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.
Physics
1 answer:
iVinArrow [24]3 years ago
5 0

Answer: 30.34m/s

Explanation:

The sum of forces in the y direction 0 = N cos 28 - μN sin28 - mg

Sum of forces in the x direction

mv²/r = N sin 28 + μN cos 28

mv²/r = N(sin 28 + μcos 28)

Thus,

mv²/r = mg [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/r = g [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]

v²/36 = 9.8 [(0.4695 + 0.87*0.8829) - (0.8829 - 0.87*0.4695)]

v²/36 = 9.8 [(0.4695 + 0.7681) / (0.8829 - 0.4085)]

v²/36 = 9.8 (1.2376/0.4744)

v²/36 = 9.8 * 2.6088

v²/36 = 25.57

v² = 920.52

v = 30.34m/s

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Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

W = F*d

Where:

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d: is the distance = 2.38 m  

Hence, the work is:

W = F*d = 108.915 N*2.38 m = 259.22 J

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!                                                

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Three collinear forces,F1=45N west,F2=63N east and an unknown force F3 are applied to an object.The resultant force of the three
Greeley [361]

Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is

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which is positive, so it's directed east.

To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that

18 N + <em>F</em>₃ = -12 N

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6 0
3 years ago
Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q
katen-ka-za [31]

Answer:

c) 4.2*10^{-5}C

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

F=\frac{kq_{1}q_{2}}{r^{2}}

where k is a proportionality constant with the value k=9*10^{9}\frac{Nm^{2}}{C^{2}}

In this case q_{1}=q_{2}=q, so we have:

F=\frac{kq^{2}}{r^{2}}

Solving the equation for q, we have:

kq^{2}=Fr^{2}

q^{2}=\frac{Fr^{2}}{k}

q=\sqrt{\frac{Fr^{2}}{k}}

Replacing the given values:

q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

q=4.2*10^{-5}C

3 0
3 years ago
The French high-speed train travels at 300 km/h. How long
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Answer:

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Explanation:

4 0
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