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Firdavs [7]
3 years ago
8

By what factor does the drag force on a car increase as it goes from 65 to 110 km/h?

Physics
2 answers:
GrogVix [38]3 years ago
5 0

Answer:

The drag force is increase by 2.9

Explanation:

The drag force on the car is:

F_{drag} =\frac{1}{2} pCAv^{2}

Where p is the density, A is the area of the car facing the air, C is the drag coefficient and v is the velocity. If k=(pCA)/2, then we have:

Fdrag=kv^2

For the speed 65 km/h:

Fdrag1=k(65^2)

For the speed 110 km/h

Fdrag2=k(110^2)

The ratio is:

Ratio=Fdrag2/Fdrag1=(k*65^2)/(k*110^2)=2.9

Art [367]3 years ago
3 0

I=2.86     HOPE I HELPED


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A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
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(a) Wavelength is 0.436 m

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As per the question:

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y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

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We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

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(a) K = \frac{2\pi}{\lambda}

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K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

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4 0
3 years ago
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