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Y_Kistochka [10]
3 years ago
5

A car with mass 1600 kg drives around a flat circular track of radius 28.0 m. The coefficient of friction between the car tires

and the track is 0.830. The car is traveling at the maximum speed possible without losing traction.
Physics
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

v=15.24 m/s

Explanation:

Given that

Mass ,m= 1600 kg

radius ,r= 28 m

Coefficient of friction ,μ = 0.83

The radial force on the car when it takes turn

F=\dfrac{mv^2}{r}

The friction force on the car

Fr= μ m g

The condition for motion without losing traction

F= Fr

\dfrac{mv^2}{r}=\mu\ m g

v²=μ r g

v=\sqrt{\mu\ r g}

Now by putting the values

v=\sqrt{0.83\times 28\times 10}\ m/s  ( take g=10m/s²)

v=15.24 m/s

The speed of the car will be 15.24 m/s

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Complete the wording of the Law of Conservation of Matter and Energy (as defined in the text).
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Answer:

Neither "matter" nor "energy" can be created or destroyed, but they can be changed from one form to the other.

Explanation:

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3 years ago
What is the rotational kinetic energy of the Earth about the Sun? Assume the earth is a uniform sphere, mass of the Earth is 5.9
Thepotemich [5.8K]

Answer:

2.66x10^33 J

Explanation:

In order to do this, we first need to know the expression for kinetic energy:

E = 1/2 I*w²  (1)

I is moment of innertia

w is angular speed.

Moment of Innertia can be calculated using the following expression:

I = 2/5 M*R²  (2)

M is mass of earth, R is radius of earth

Replacing the data in expression (2) we have:

I = 2/5 * 5.97x10^24 * (6.37x10^6)²

I = 9.69x10^37 kg m²

Next, we need to calculate the angular speed of Earth over it's axis, this is easy, because we know the Earth rotates over it's own axis once a day, 24 hours (86400 s), and assuming Earth is a perfect sphere, we can calculate the speed:

w = 2π / 86400 = 7.27x10^-5 rad/s

next thing we need to do is calculate the rotational kinetic energy of earth on it's axis, using equation (1) so:

E = 1/2 * 9.69x10^37 * (7.27x10^-5)²

E = 2.56x10^29 J

Now that we have this value, we can finally calculate the rotational kinetic energy of earth about the sun. For that, we need to calculate again the angular speed of earth about the sun. The Earth rotates around the sun once a year, or 365 days, which is 3.1536x10^7 s, so the angular speed would be:

w = 2π/3.1536x10^7 = 1.99x10^-7 rad/s

finally the energy is the combination of the sun and earth so:

K = 1/2 (Ie + Me*Rorb²)wo²

I is innertia for earth

Me mass of earth

Rorb RAdius of orbit around the sun

wo is angular speed around the sun

Replacing the data we finally have:

K = 1/2 [9.69x10^37 + 5.97x10^24 * (1.5x10^11)²]*(1.99x10^-7)²

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4 0
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Answer:

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Explanation:

Sine they have Same distance D

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D = 8.3312(t+55.9)

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D=5.08 t

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