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emmasim [6.3K]
3 years ago
7

A kite suspended in the sky is flowing back and forth. Which type of friction is being described?

Physics
2 answers:
solniwko [45]3 years ago
7 0
It would be fluid friction.
PilotLPTM [1.2K]3 years ago
7 0

Answer:

Fluid friction - in fluid friction, there is something called air resistance. This makes fluid friction happen in liquids AND in gases.

Explanation:

You might be interested in
Light travels 186,282 miles in one second. A light year is the distance light travels in one year. It is about 4.3 light years t
deff fn [24]

Answer:

2.5278035226\times 10^{13}\ miles

1994.7944465 miles

46390568.5227 miles

125 ft

Explanation:

1\ ly=186282\times 365.25\times 24\times 60\times 60=5.8786128432\times 10^{12}\ mi

4.3\ ly=4.3\times 5.8786128432\times 10^{12}=2.5278035226\times 10^{13}\ miles

The distance is 2.5278035226\times 10^{13}\ miles

2000\ miles=0.01\ in

2.5278035226\times 10^{13}\times \dfrac{1}{2000}=1.2639017613\times 10^{10}\ LA\ to\ NY

Converting to inches

1.2639017613\times 10^{10}\times 0.01=126390176.13\ in/LA\ to\ NY

Converting to miles

126390176.13\times\dfrac{1}{5280\times 12}=1994.7944465\ inches/LA\ to\ NY

The distance is 1994.7944465 miles

5.8786128432\times 10^{17}\times \dfrac{1}{2000}\times 0.01\times\dfrac{1}{5280\times 12}=46390568.5227\ miles

The distance is 46390568.5227 miles

5.8786128432\times 10^{12}\times 15\times 10^{9}=8.8179192648\times 10^{22}

8.8179192648\times 10^{22}\times \dfrac{1}{5.8786128432\times 10^{17}}=150000\ MW\ dia

150000\dfrac{1}{100\times 12}=125\ ft

The distance is 125 ft

5 0
4 years ago
A wheel accelerates so that it's angular speed increases uniformly from 150 rads/s to 580 rads/s in 16 revolutions.Cakcjlate its
LenaWriter [7]

Answer:

A = 26.875 rad/s²

Explanation:

<u>Given the following data;</u>

Initial angular speed, Uw = 150 rads/s.

Final angular speed, Vw = 580 rads/s.

Time = 16 seconds.

To calculate the angular acceleration;

From kinematics equation;

At = Vw - Uw

Where;

  • A is the angular acceleration.
  • t is the time
  • Vw is the final angular speed.
  • Uw is the initial angular speed.

Substituting into the formula, we have;

A*16 = 580 - 150

16A = 430

A = 430/16

<em>A = 26.875 rad/s²</em>

3 0
3 years ago
If a constant fotce of 490 newtons is applied to an object and it proceeds to accelerate in the opposjte direction at 9.8 meters
Sloan [31]

Answer:

50 kg

Explanation:

f=ma

Force = mass x acceleration

490 = 9.8 x m

m = 50 kg

8 0
3 years ago
The force stretching the D string on a certain guitar is 130 N. The string's linear mass density is 0.006 kg/m. What is the spee
inysia [295]

Answer:

The value is  v  = 147.2 \  m/s

Explanation:

From the question we are told that

    The force is   F =  130 \  N

     The linear mass density is  \mu =  0.006 \  kg/m

     

Generally the speed of the wave on the string is mathematically represented as  

            v  = \sqrt{\frac{F}{\mu } }

=>        v  = \sqrt{\frac{130}{0.006} }

=>        v  = 147.2 \  m/s

3 0
3 years ago
Una rueda que gira a 300 r.P.M. Aumenta su velocidad bajo una aceleración angular de 6 2 s rad : calcula: a) La velocidad angula
Rina8888 [55]

Answer:

a)   w = 873 rev,  b)  θ = 97.75 rev

Explanation:

This is a rotation kinematics exercise

         w = w₀ + α t

         θ = θ₀ + w₀ t + ½ α t²

let's start by reducing the magnitudes to the SI system

         w₀ = 300 rpm (2pi rad / 1 rev) (1 min / 60s) = 31.42 rad / s

          α = 6 rad / s²

a) let's look for the angular velocity

            w = 31.42 + 6 10

             w = 91.42 rad / s

b) θ₀ = 0

             θ = 0 + 31.42 to + ½ 6 10²

             θ = 614.2 rad

As they ask for the result in rpm and revolutions, let's carry out the reduction

         w = 91.42 rad / s (1 rev / 2pi rad) (60 s / 1min)

         w = 873 rev

         θ = 614.2 rad (1 rev / 2pi rad)

         θ = 97.75 rev

8 0
3 years ago
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