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kenny6666 [7]
4 years ago
6

car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. 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 the track
Physics
1 answer:
Luba_88 [7]4 years ago
7 0

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

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Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
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Answer:

19 m/s

Explanation:

The complete question requires the final speed to be calculated.

Velocity is the rate and direction at which an object moves. Acceleration is the rate of change of velocity per unit time and can be calculated by the difference in velocity over a given time.

For this question, first the unknown acceleration must be calculated and used to determine the final velocity

Step 1: Calculate the acceleration

a=\frac{v_{2}-v{1}}{t_{1}}

a=\frac{11-7}{8}

a=\frac{4}{8}

a=0.5 m/s^{2}

Step 2: Calculate the velocity using the acceleration calculated above

a=\frac{v_{3}-v{2}}{t_{2}}

0.5=\frac{v_{3}-11}{16}

v_{3}=19 m/s

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Based on Kepler's work, which best describes the orbit If a planet around the Sun?
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an ellipse with the Sun at one focus  or D

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In a 120-volt circuit having a resistance of 12 ohms, the power W in watts when a current I is flowing through is given by W=120
Mrac [35]

Answer:

maximum power = 300 W

Explanation:

Ohm's law: Ohm's law state that the current flowing through a metallic conductor, is directly proportional to the potential difference applied across its end. mathematically it is expressed as,

V = IR............. Equation 1

Where V = potential difference, I = current, R = Resistance of the conductor.

If the power flowing through is gives as

W = 120I - 12I² ..................... Equation 2

To get the maximum power we differentiate of equation 2  and equate to zero

dW/dt = 0

120 - 24l = 0........................... equation 3

Making I the subject of the equation,

I = 120/24 = 5 A.

Suubstituting the value of I into Equation 2

W = 120 (5) - 12(5)²

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W = 300 W.

Therefore maximum power = 300 W

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4 years ago
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