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Papessa [141]
3 years ago
9

Calculate the terminal velocity of a rain drop of radius 0.12cm​

Physics
1 answer:
Tasya [4]3 years ago
6 0

Explanation:

Given that,

The radius of rain drop, r = 0.12 cm = 0.0012 m

The viscocity of air is, \eta=18\times 10^{-5}\ poise

Let the viscous force is, F = 0.010173\ N

The viscous force is given by :

F=6\pi \eta rv\\\\v=\dfrac{F}{6\pi \eta r}

Put all the values,

v=\dfrac{0.010173}{6\pi 18\times 10^{-5}\times 0.0012 }\\\\v=2498.58\ m/s

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How to cure bor-edom????
Allisa [31]

Answer:

keep ur self busy

Explanation:

read a book

catch up on homework

get ahead of the class

watch tv

go outside

4 0
3 years ago
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7) the observation deck of the Empire State Building is 381 m above the street. Determine the
Elan Coil [88]

Answer:

Questions involving accelerating objects, including those in free fall, can be solved using three equations

 

y =  1/2at2 + v0t + y0

 

v = v0 + at

 

vf2 = v02 + 2 a (y - y0)

 

In this case, only the first equation is needed.

 

Since the object is falling the final height is 0 meters, the initial height is 370 m, and the acceleration is -9.8ms-2

 

Since the object is dropped, the initial velocity is 0.

 

0 = 1/2(-9.8)t2 + 370.    t = sqrt(370/4.9) = 8.69 s

Explanation:

8 0
4 years ago
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A parallel-plate capacitor with circular plates of radius R is being discharged. The displacement current through a central circ
Dvinal [7]

Answer:

The discharging current is I_d  =  36.8 \ A

Explanation:

From the question we are told that  

     The radius of each circular plates is  R

     The displacement current is  I  = 9.2 \ A

      The radius of the central circular area is  \frac{R}{2}

The discharging current is mathematically represented as

       I_d  =  \frac{A}{k} *  I

where A is the area of each plate which is mathematically represented as

       A  =  \pi R ^2

and   k is central circular area which is mathematically represented as

     k  =  \pi [\frac{R}{2} ]^2

So  

     I_d  =  \frac{\pi R^2 }{\pi * [ \frac{R}{2}]^2 } *  I

     I_d  =  \frac{\pi R^2 }{\pi *  \frac{R^2}{4} } *  I

     I_d  =  4 *  I

substituting values

     I_d  =  4 *  9.2

     I_d  =  36.8 \ A

     

7 0
3 years ago
Calculate the electrostatic force that a small sphere A, possessing a net charge of 2.0 x 10^-6 Coulombs exerts on another small
Anni [7]

Answer:

F = 5.33*10^-4N

Explanation:

to find the electrostatic force you use the Coulomb's law, given by the formula:

F=k\frac{q_Aq_B}{r^2}

k: Coulomb's constant = 8.89*10^9 Nm^2/C^2

q_a: charge of A = 2.0*10^{-6}C

q_B: charge of B = -3.0*10^{-6}C

r: distance between the spheres = 10.0m

By replacing all these values you obtain:

F=(8.89*10^9Nm^2/C^2)\frac{(2.0*10^{-6}C)(-3.0*10^{-6}C)}{(10.0m)^2}=5.33*10^{-4}N

hence, the forcebetween the spheres is about 5.33*10^-4N

3 0
3 years ago
A listener is sitting somewhere on the line between two loudspeakers that are 10 m apart. The speakers are each emitting a sine
prohojiy [21]

Answer:

57.17 Hz 114.34 Hz 285 Hz

Explanation:

The distance between the men and 1 speaker = 3.5 m

Distance between the men and second speaker = 10-3.5= 6.5 m

Here at this point there will be no sound so there will be destructive interference

Path difference \Delta x=6.5-3.5=3

We know that for destructive interference \Delta x=(2m+1)\frac{\lambda }{2}=(2m+1)\frac{v}{2f}

3=(2m+1)\frac{v}{2f}

f=(2m+1)\frac{v}{6} here v is the speed of sound in air

So for m =0

f=(2\times 0+1)\frac{343}{6}=57.17H

for m =1

f=(2\times 1+1)\frac{343}{6}=114.34Hz

for m=2

f=(2\times 2+1)\frac{343}{6}=285Hz

 

6 0
3 years ago
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