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Digiron [165]
3 years ago
7

A race car travels a circular track at an average rate of 135 mi/hr. The radius of the track is 0.450 miles. What is the centrip

etal acceleration of the car?
Physics
2 answers:
kenny6666 [7]3 years ago
7 0
135 * 135 / 0.450 = 40,500 mi/hr2 

PilotLPTM [1.2K]3 years ago
4 0

Answer : a_c=40500\ mi/h^2.

Explanation :

It is given that,

Velocity of a race car, v = 135 mi/hr

Radius of the track, r = 0.45 miles

We know that the centripetal acceleration is defined as :

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

r is the radius of the circle.

a_c=\dfrac{(135\ mi/hr)^2}{0.45\ mi}

a_c=40500\ mi/h^2

So, the centripetal acceleration of the car is a_c=40500\ mi/h^2.

Hence, this is the required solution.

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<u>for</u><u> </u><u>the</u><u> </u><u>second</u><u> </u><u>one</u><u> </u><u>:</u>

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4 years ago
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
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Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

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Using the magnification formula

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4) If u = 15cm

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5) if u = 20cm

1/v = 1/f-1/u

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1/v = 2-1/20

1/v = 1/20

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20H1 = 100

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6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

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