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DENIUS [597]
3 years ago
10

I really need help with this. please! thank you. ​

Physics
1 answer:
zhannawk [14.2K]3 years ago
7 0
Solo estoy aquí por la libra, lo siento, no pude ayudar
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Give two examples of where a permanent magnet might used. <3 
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Two examples of where you can find permanent magnets are
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2.speakers
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Which molecule will form a solution with a very low pH?
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The correct answer is A: H3PO4
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Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vert
chubhunter [2.5K]

Answer:

362.41 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.8

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.04 m²

v = Terminal velocity

F = ma

F=\frac{1}{2}\rho CAv^2\\\Rightarrow mg=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{mg}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{20\times 9.81}{1.21\times 0.8\times 0.04}}\\\Rightarrow v=100.66924\ m/s

Converting to km/h

100.66924\times 3.6=362.41\ km/h

The terminal velocity of the stone is 362.41 km/h

5 0
3 years ago
Read 2 more answers
A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

  • <u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

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3 years ago
Electrical power companies sell electrical energy
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Heat used by electric heater :

Q = m • c • ∆T

Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)

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Cost of electrical energy :

Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)

Cost = $ 0.3675

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