COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 5, Problem 84QAP
To determine

(a)

The value the drag constant c.

Expert Solution
Check Mark

Answer to Problem 84QAP

The value the drag constant c is 4.54×106kg/m.

Explanation of Solution

Given:

Terminal velocity of the raindrop

  v=8.50 m/s

Diameter of the drop

  2r=4.00 mm

Density of water

  ρ=1.00×103kg/m3

Equation for the drag force

  Fdrag=cv2

Formula used:

If the drop moves with terminal velocity, then the drag force acting upwards is equal to the drop's weight w.

  Fdrag=w

If m is the mass of the drop, V its volume and g the acceleration of free fall, then,

  w=mg

And,

  m=ρV

Since the volume is given by,

  V=43πr3, therefore,

  m=ρV=43πr3ρ

This follows that,

  Fdrag=43πr3ρg

Since Fdrag=cv2,

  cv2=43πr3ρg

Write an expression for c.

  c=4πr3ρg3v2......(1)

Calculation:

Determine the radius of the drop and express it in m.

  r=2r2=4.00 mm2=2.00 mm×103m1 mm=2.00×103m

Substitute the given values in the expression and calculate the value of c.

  c=4πr3ρg3v2=4(3.14)( 2.00× 10 3 m)3(1.00× 10 3 kg/m 3)(9.80  m/s 2)3( 8.50 m/s)2=4.54×106kg/m

Conclusion:

The value the drag constant c is 4.54×106kg/m.

To determine

(b)

The value of the terminal velocity of a drop of diameter 8.00 mm under the same conditions.

Expert Solution
Check Mark

Answer to Problem 84QAP

The value of the terminal velocity of a drop of diameter 8.00 mm

is 24.04 m/s.

Explanation of Solution

Given:

The terminal velocity of the drop of diameter 4.00 mm

  v=8.50 m/s

Diameter of the drop 1

  2r=4.00 mm

Diameter of the drop 2

  2r'=8.00 mm

Formula used:

From the equation

  c=4πr3ρg3v2,

it can be seen that, since

  v=43cπr3ρg

Therefore,

  vr3

Therefore,

  v'v=r'3r3......(2)

Calculation:

The diameter of the drop 2 is 8.00 mm, which is 2 times the diameter of drop 1.

Therefore,

  r'=2r

The equation (2) reduces to,

  v'v=r ' 3 r 3=( 2r r )3=8

Substitute the value of the terminal velocity of drop1 and calculate the value of the terminal velocity of drop 2.

  v'=8v=2.82(8.50 m/s)=24.04 m/s

Conclusion:

The value of the terminal velocity of a drop of diameter 8.00 mm

is 24.04 m/s.

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Chapter 5 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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