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aev [14]
3 years ago
14

If I hit a guy in the mouth with my left hand even though I'm right handed, would it count as me hitting them?

Physics
1 answer:
Lera25 [3.4K]3 years ago
3 0

Answer:

yes it would u just wouldn't be able to hit him as hard as u would if u were using your dominate hand

Explanation:

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An arrow is launched from a bow with an initial horizontal velocity of 40
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V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2

V = 73.8 m/s

tan theta = Vy / Vx = 62/40 = 1.55

theta = 57.2 deg

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5. Sarah is standing 32.0 m away from Jason. If Sarah throws a snowball at a 45 degree
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Answer:

Explanation:

If air resistance is ignored and assume UP and Toward Jason are the positive directions.

horizontal analysis

d = (vx₀)t

t = d/vx₀

horizontal analysis

0 = vy₀t + ½gt²

0 = vy₀(d/vx₀)+ ½g(d/vx₀)²

as vy₀ = v₀sin45 and vx₀ = v₀cos45 and are equal.

0 = d + ½g(d²/v₀²cos²45)

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   v₀² = -½g(d/cos²45)

   v₀² = -½(-9.81(32.0/cos²45)

   v₀² = 313.92

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7 0
3 years ago
WGVU-AM is a radio station that serves the Grand Rapids, Michigan area. The main broadcast frequency is 1480kHz. At a certain di
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Answer:

a

  \lambda  =  202.7 \  m

b

  w =  9.3 *10^{6} \  rad/s

c

  k = 0.031 m^{-1}

d

  E_{max} = 9.0 *10^{-3} \  V/m

Explanation:

From the question we are told that  

       The frequency of the radio station is  f=  1480 \  kHz  =  1480 *10^{3}\ Hz

         The magnitude of the magnetic field is  B  =  3.0* 10^{-11} \  T

Generally the wavelength is mathematically represented as

          \lambda  =  \frac{c}{f}

Here c is the speed of light with value  c =  3.0 *10^{8} \ m/s

So

        \lambda  =  \frac{3.0 *10^{8}}{ 1480 *10^{3}}

=>      \lambda  =  202.7 \  m

Generally the angular frequency is mathematically represented as

       w =  2 \pi * f

=>   w =  2 * 3.142 *  1480 *10^{3}

=>   w =  9.3 *10^{6} \  rad/s

Generally the wave number is mathematically represented as        

=>      k = \frac{2 \pi }{\lambda}

=>      k = \frac{2  *  3.142  }{ 202.7}

=>      k = 0.031 m^{-1}

Generally the amplitude of the electric field at this distance from the transmitter is mathematically represented as

         E_{max} = c *  B

=>      E_{max} = 3.0 *10^{8} *   3.0* 10^{-11}

=>      E_{max} = 9.0 *10^{-3} \  V/m

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