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skad [1K]
2 years ago
14

Can a 20 N force and 40 N force ever produce a resultant with magnitude of 27 N?

Physics
1 answer:
SVEN [57.7K]2 years ago
7 0

Sure ,Let's find angle between forces

  • Vectors be A and B and resultant be R

\\ \sf\longmapsto R^2=A^2+B^2+2ABcos\theta

\\ \sf\longmapsto 27^2=20^2+40^2+2(20)(40)cos\theta

\\ \sf\longmapsto 729=400+1600+1600cos\theta

\\ \sf\longmapsto 729=2000+1600cos\theta

\\ \sf\longmapsto 1600cos\theta=-271

\\ \sf\longmapsto cos\theta=-0.169

\\ \sf\longmapsto \theta=cos^{-1}(-0.169)

\\ \sf\longmapsto \theta=80.2°

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Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio
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The gravitational potential energy of the brick is 25.6 J

Explanation:

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3 years ago
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A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity
nadezda [96]

Answer:

u(t)=\frac{1}{5} sin\ (25t)

Explanation:

Given:

  • mass of the body stretching the spring, m=150\ g
  • extension in spring, \Delta x=1.568\ cm
  • velocity of oscillation, u'(0)=20\ cm.s^{-1}
  • initial displacement position of equilibrium, u(0)=0

<u>According to given:</u>

m.g=k.\Delta x

150\times 980=k\times 1.568

k=93750\ dyne.cm^{-1}

<u>we know frequency:</u>

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{93750}{150} }

\omega=25

Now, for position of mass in oscillation:

u= A.sin\ (\omega.t)+B.cos\ (\omega.t)

u= A.sin\ (25.t)+B.cos\ (25.t)

at t=0;\ u(0)=0\ \Rightarrow A=0

∴u(t)=B.sin\ (25.t)

∵ at t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}

u(t)=\frac{1}{5} sin\ (25t)

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