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kenny6666 [7]
3 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]3 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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A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
3 years ago
During science class, while studying mixtures, you mix together iron filings and sand. Your teacher challenges you to separate t
Lapatulllka [165]

Answer : Use a magnet to pull out the iron filings as they are attracted to a magnet

Explanation :  we use the magnet to separate the sand from the iron filing. Because the magnet has an attraction power.

We can say that we can use the magnet to attract the iron filings out of the mixture because iron is magnetic solid, but sand will not attract because sand is not magnetic solid.

So,  we use a magnet to pull out the iron filings as they are attracted to a magnet.

8 0
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Explanation:

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Mass = (120 N) / (9.8 m/s²)

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How are changes in energy and matter related
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Answer:

According to Einstein's famous equation, matter can convert into energy (and viceversa) as follows:

E=mc^2

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E is the energy

m is the mass

c is the speed of light (3\cdot 10^8 m/s)

Given the huge value of c^2, we see that even a tiny amount of matter is able to release a huge amount of energy, when the whole mass is converted into energy. This is precisely what happens in nuclear reactions. For example, in the process of nuclear fusion (that occurs in the core of the stars), two light nuclei fuse together into a heavier nucleus. The mass of the final nucleus is lower than the total mass of the initial nuclei, so part of the mass has been converted into energy according to the equation above: this is why the amount of energy produced by stars is so big.

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