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alexira [117]
3 years ago
10

Formula One race cars are capable of remarkable accelerations when speeding up, slowing down, and turning corners. At one track,

cars round a corner that is a segment of a circle of radius 95 m at a speed of 68 m/s. What is the approximate magnitude of the centripetal acceleration, in units of g?
Physics
1 answer:
anygoal [31]3 years ago
3 0

Answer:

Centripetal acceleration of the car is (4.96 g) m/s²

Explanation:

It is given that,

Radius of circle, r = 95 m

Speed of the car, v = 68 m/s

We need to find the centripetal acceleration. It is given by :

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

So, a_c=\dfrac{(68\ m/s)^2}{95\ m}

a_c=48.67\ m/s^2

Since, g = 9.8 m/s²

So,

a_c=(4.96\ g)\ m/s^2

So, the magnitude of the centripetal acceleration is (4.96 g) m/s². Hence, this is the required solution.

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The distance from a 27 mw point source of electromagnetic waves where the electric field amplitude 0. 060 v/m will be 21.21 m .

Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. In other words, EM waves are composed of oscillating magnetic and electric fields.

The highest point of a wave is known as 'crest' , whereas the lowest point is known as 'trough'. Electromagnetic waves can be split into a range of frequencies. This is known as the electromagnetic spectrum.

c = 3 * 10^{8} m/s

∈ = 8.85 * 10^{- 12} C^{2} / N/ m^{2}

E = 0. 060 v/m

I = P / 4πr^{2}

Also , I = c ∈ E^{2} /2

r^{2} = P / 4π I                                 equation 1

substituting the value of I in equation 1

r^{2} = 2 P / 4π (c ∈ E^{2} )

r^{2} = 2 * 27 * 10^{-3} / 4 * 3.14 * 3 * 10^{8} * 8.85 * 10^{- 12}  * (0.06)^{2}

r = 21.21 m

To learn more about Electromagnetic waves  here

brainly.com/question/12392559

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