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il63 [147K]
3 years ago
14

What is the maximum speed with which a 1200-kg car can round a turn of radius 93.0 m on a flat road if the coefficient of static

friction between tires and road is 0.60
Physics
2 answers:
gulaghasi [49]3 years ago
8 0

To solve this problem, apply the concepts related to the centripetal force expressed as the product between the mass and the speed squared over the radius of gyration, and the concept of the friction force equivalent to the product between the coefficient of friction, the mass of the body and acceleration due to gravity, for the problem we equate the centripetal force with the frictional force we have

F_c = F_f

\frac{mv^2}{r} = \mu mg

Rearrange the expression to get the value of speed

v = \sqrt{r\mu g}

Replacing with our values we have,

v = \sqrt{(93m)(0.60)(9.8m/s^2)}

v = 23.38m/s

Therefore the velocity is 23.38m/s and we can conclude also that the maximum speed of the car is independent of the mass of the car.

Mashcka [7]3 years ago
7 0

Answer:

The maximum speed is 23.39m/s

Explanation:

Centripetal force Fc is the force acting on a body moving in a circular path and directed towards the centre of the path.

If F = ma

centripetal acceleration a = v²/r

Centripetal force Fc = mv²/r ... (1)

where;

m is the mass of the object

r is the radius.

Since there is friction between the tyre and the road, then the frictional force Ff acts between the surface and this frictional force is the one that tends to oppose the moving force (centripetal force)

Ff = μsR where

μs is the coefficient of static friction

R is the normal reaction which is also equivalent to the weight of the car i.e R = W = mg

Ff = μsmg ... (2)

For the body to be static, the centripetal force must be equal to the frictional force i.e Fc = Ff

mv²/r = μsmg

Making v the subject of the formula;

v²/r = μsg

v² = μsgr

v = √μsgr

Given the following data;

μs = 0.6

g = 9.81m/s²

r = 93.0m

v = √0.6×9.81×93

v = √547.398

v = 23.39m/s

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SashulF [63]

Answer:

 L = 41.09 Kg m2 / s      The angular momentum does not depend on the time

Explanation:

The definition of angular momentum is

        L = r x p

Where blacks indicate vectors

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Let's replace

      L = m r x v

The given function is

      x = 6.00 i ^ + 4.15 t j ^

We look for speed

     v = dx / dt

     v = 0 + 4.15 j ^

To evaluate the angular momentum one of the best ways is to use determinants

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7 0
3 years ago
an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco
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Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

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a_c = \frac{v^2}{r}

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ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

a_c = \frac{(12\ m/s)^2}{2\ m}

<u>ac = 72 m/s²</u>

<u></u>

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F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)

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6 0
2 years ago
A charged particle is located in an electric field where the magnitude of the electric field strength is 2.0x10^3 newtons per co
Alexxandr [17]

Answer:

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6 0
3 years ago
Learning Goal:
enot [183]

Answer:

A. U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

C=\dfrac{\epsilon A}{d}

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For air or free space, \epsilon is \epsilon_0 called the permittivity of free space. In general, \epsilon=\epsilon_r \epsilon_0 where \epsilon_r is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, \epsilon_r=1.

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Its charge, Q, is related to its capacitance by Q=CV (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for U,

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B. When the distance is 3d,

U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}

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6 0
3 years ago
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Radda [10]

Answer:

Hey shaikaadil700 !

<u> </u><u>Lubricating</u><u> </u> of rough surfaces reduces friction.

Explanation:

• Lubricating is the smoothening or polishing of the surfaces

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3 0
3 years ago
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