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Wewaii [24]
2 years ago
10

A small water droplet in a mist of air is approximated as being a sphere of diameter 1.25 mil. Calculate the terminal velocity a

s it falls through still air to the ground.
Physics
1 answer:
shtirl [24]2 years ago
8 0

The terminal velocity as it falls through still air is 4.65154 in/s.

The diameter of small water droplet is 1.25 mil= 1.25×0.0254×10^-3 m

                                                                            = 3.175 × 10^-5 m

Now the viscosity of still air is η = 1.83× 10⁻⁵ Pa

 So the formula for drag force is:

                Fd = 6πηrv

  where,  v is the velocity.

Now to attain terminal velocity acceleration must be zero.

            →  W = Fd

                ρVg = 6πrηv

                ρ × 4/3 πr³×g = 6πrηv

                           v = 2/9 × ρgr³/ η

                           v =  2/9 × 10³×9.81×(3.175×10^-3) / 18.6×10^-6

                           v = 0.1181  m/s

                           v = 4.65154 in/s

Learn more about terminal velocity here:

     brainly.com/question/20409472

              #SPJ4

                   

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A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
sladkih [1.3K]

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

7 0
3 years ago
An ideal gas occupies 600 cm3 at 20c. at what temperature will it occupy 1200 cm3 if the pressure remains constant? 10c 40c 100c
Anna11 [10]
ANS : 313℃
You need to use K in this.
To convert​ ℃ to Kelvin (K), add 273.15 to ℃.

5 0
4 years ago
"5 N, up" is an example of a ___.<br> OA) force<br> OB) mass<br> OC) weight<br> OD) magnitude
Sunny_sXe [5.5K]

Answer:

A) Force

Explanation:

It is an example of force since force is a vector quantity so it has magnitude and direction. In this case the magnitude is equal to 5 [N] and the direction is upward.

The weight can not be, as it always acts downward.

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5 0
3 years ago
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Pavlova-9 [17]

Answer:

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Explanation:

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The battery rating is 170 Ampere-hours, therefore:

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But 1 hour = 3600 seconds;

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The total charge that the battery can provide is 612000 C.

8 0
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Schach [20]
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