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a_sh-v [17]
3 years ago
8

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was

μs = 1. around 1962, three companies independently developed racing tires with coefficients of 1.6. this problem shows that tires have improved further since then. the shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. ]
Physics
1 answer:
True [87]3 years ago
5 0

This problem is looking for the minimum value of μs that is necessary to achieve the record time. To solve this problem:


Assuming the front wheels are off the ground for the entire ¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².


For a constant acceleration, distance = 402.3


m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2



µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18

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In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten
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Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

V = iR

V = (2 mA)(10 kohm)

V = 20 volts

so voltage across the capacitor + voltage across resistor = V

V_c + 20 = 50

V_c = 30 V

Now we know that

U = \frac{q^2}{2C}

here rate of change in energy of the capacitor is given as

\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}

\frac{dU}{dt} = (30)(2 mA)

\frac{dU}{dt} = 0.06 W

3 0
3 years ago
A 5 kg block is sliding down a plane inclined at 30^0 with a constant velocity of 4 m/s. To determine the coefficient of frictio
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Answer:

The necessary information is if the forces acting on the block are in equilibrium

The coefficient of friction is 0.577

Explanation:

Where the forces acting on the object are in equilibrium, we have;

At constant velocity, the net force acting on the particle = 0

However, the frictional force is then given as

F = mg sinθ

Where:

m = Mass of the block

g = Acceleration due to gravity and

θ = Angle of inclination of the slope

F = 5×9.81×sin 30 = 24.525 N

Therefore, the coefficient of friction is given as

24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479

μ × 42.479 N= 24.525 N

∴ μ = 24.525 N ÷ 42.479 N = 0.577

3 0
3 years ago
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4 0
3 years ago
4. What is the magnitude and direction of the gravitational force that acts on a man
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The gravitational force acting on the man is 800 N towards the Earth's centre

Explanation:

The weight of an object on the Earth is exactly the gravitational force exerted by the Earth on the object.

The gravitational force exerted by the Earth on an object located at the Earth's surface is given by:

F=G\frac{Mm}{R^2}

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

R is the radius of the Earth

And the direction of the force is towards the Earth's centre.

Since G, M and R are constant, they are grouped into a single constant called g, acceleration of gravity:

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therefore the gravitational force can be rewritten as

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And this is the usual equation that we use to calculate the weight of an object.

Therefore, weight and gravitational force acting on an object on Earth are the same thing: so, the gravitational force acting on the man is equal to his weight, 800 N, and it acts towards the Earth's center.

Learn more about weight and gravitational force:

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brainly.com/question/12785992

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