Physics, 18.01.2021 14:00 puppycity78

Heather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The Super-Beetle has a mass of 1500 kg and the coefficient of static friction between the road and tires is 0.6. a. What is the force of static friction the road can apply batore the car starts to selon (use Ft= uFn).
b. What is the maximum speed the car can travel before it would start to slide?

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Answer from: michaelmontoya64

a) The force of static friction the road can apply before the car starts to move is 8826.3 newtons.

b) The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.

Explanation:

a) Let suppose that the car is on a horizontal ground and travels at constant speed. The vehicle experiments a centripetal acceleration due to friction, which can be seen in the Free Body Diagram (please see image attached for further details). By Newton's Laws, we construct the following equations of equilibrium:

$\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (1)

$\Sigma F_{y} = N -m\cdot g = 0$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

- Normal force from ground to the car, measured in newtons.

- Maximum speed of the car, measured in meters per second.

- Radius of the circular track, measured in meters.

- Mass, measured in kilograms.

- Gravitational acceleration, measured in meters per square second.

By applying (2) in (1):

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$ (3)

The force of static friction the road can apply in the car (), measured in newtons, is: ( $\mu_{s} = 0.6$ , $m = 1500\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ )

$f = \mu_{s}\cdot m \cdot g$

$f = (0.6)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$f = 8826.3\,N$

The force of static friction the road can apply before the car starts to move is 8826.3 newtons.

b) Then, we calculate the maximum speed of the car by (3):

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$

$\mu_{s}\cdot g = \frac{v^{2}}{R}$

$v = \sqrt{\mu_{s}\cdot g\cdot R}$

If we know that $\mu_{s} = 0.6$ , $g = 9.807\,\frac{m}{s^{2}}$ and $R = 200\,m$ , then the maximum speed of the car can travel before it would start to slide is:

$v =\sqrt{(0.6)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (200\,m)}$

$v \approx 34.305\,\frac{m}{s}$

The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.

Heather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The S

Answer from: Quest

what is the actual question though?

Answer from: Quest

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

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