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Burka [1]
3 years ago
9

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the magnitude of the

drag force? a.The drag force goes up by a factor of 4 b.The drag force stays the same. c.The drag force decreases. d.The drag force doubles as well
Physics
1 answer:
SpyIntel [72]3 years ago
5 0

Answer:

F_d,1 / F_d,o =  4 , Option A

Explanation:

Given:

- The velocity of the bike V_o is doubled to 2*V_o

Find:

what happens to the magnitude of the drag force?

Solution:

- The expression of the Drag force as a function of objects area, coefficient of drag, velocity and density of medium is given by:

                                 F_d = 0.5*C_d*A*p*V^2 / 2

- In our case the force of drag with speed V_o is:

                                F_d,o = 0.5*C_d*A*p*V_o^2 / 2

- force of drag with speed 2V_o is:

                                 F_d,1 = 2*C_d*A*p*V_o^2 / 2

- Take the ratio of F_d,1 to F_d,o:

                  F_d,1 / F_d,o =  (2*C_d*A*p*V_o^2 / 2) / (0.5*C_d*A*p*V_o^2 / 2)

We get:                      F_d,1 / F_d,o =  4

Hence, by doubling the speed the magnitude of drag force is increased by a factor of 4.    

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Answer:

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\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}

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Explanation:

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A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
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Answer:

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Explanation:

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y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

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Igoryamba

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

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  • Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
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b)

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c)

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