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GREYUIT [131]
3 years ago
8

Consider a small raindrop and a large raindrop falling through the atmosphere. (a) Compare their terminal speeds. Both raindrops

have the same terminal speed. The larger drop has lower terminal speed. The larger drop has higher terminal speed. Correct: Your answer is correct. (b) What are their accelerations when they reach terminal speed?
Physics
1 answer:
Ivanshal [37]3 years ago
4 0

Terminal velocity is the maximum velocity attainable by an object as it falls through a fluid and is given by,

v=\sqrt{\frac{2mg}{\rho A C}}

Where,

m = mass of the falling object.

g = the acceleration due to gravity.

\rho = the density of the fluid the object is falling through.

A = the projected area of the object.

C = the drag coefficient.

A) Because a drop is difficult to approximate to a certain form, it usually tends to be considered a spectrum, the terminal velocity for a sphere is given by

V_t = \sqrt{\frac{4 g D (\rho_p-\rho)}{3 C \rho}}

Whatever our appreciations, all the variables are constant, except for the Diameter, we can realize that the terminal velocity is proportional to the radius of the object, the greater the radius - the larger the drop - the greater the terminal velocity.

B) Since there is a "constant" terminal velocity at the end of the path, at which point the forces are balanced, the acceleration will be 0. For both objects.

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xmax = 9.5cm

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You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

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