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  1. I assume that "10 cm size" means a perimeter of 10 cm . The internal angles of...1*cos108º = 2.618 cm² by the law of cosines c = 1.618 cm so the semiperimeter is s = ½(1.618 + 1 + 1) cm = 1.809 cm ...

    2 Answers · Science & Mathematics · 18/09/2020

  2. displacement = 20 cm - (- 20cm) + (- 10 cm ) = 30 cm distance covered = |(-20cm) - (20cm)| + |(-10cm) - (-20 cm )| = 40cm + 10cm = 50 cm

    1 Answers · Science & Mathematics · 05/07/2020

  3. a) at 14 cm , the "enclosed" charge is Q = 50nC * (14/30... * 5.08e-9C / (0.14m)² = 2331 N/C (outward) b) at 37 cm , the enclosed charge is 50 nC: E = 8.99e9N·m²...

    1 Answers · Science & Mathematics · 30/09/2020

  4. ...7=1.8097x 10^-5T In the smaller loop, the radius = r1 = 12.5 cm = 0.125 m I1 = 3.6 Amp hence magneti field at the center = B1...

    1 Answers · Science & Mathematics · 14/05/2020

  5. ...)/(u₁+u₂) = (6.5*22 + 14*(-15))/(6.5 + 14) = -3.27 cm /s Initial vel. of 6.5g object in CoMF = 22 - (-3.27) = 25.27 cm /s...

    1 Answers · Science & Mathematics · 18/06/2020

  6. A=4.14 cm V=3.59 V=f*λ V/λ=(A*2*pi*f)/(f*I) 3.59=(2*A*pi)/I I=(2*4.14*pi)/3.59 I=7.24

    1 Answers · Science & Mathematics · 08/07/2020

  7. ...v 1 / v = 1 / f - 1 / u 1 / v = 1 / 38 - 1 / 19 = (1 - 2) / 38 = - 1 / 38 v = - 38 cm The image distance is negative —> the image is virtual, ...

    1 Answers · Science & Mathematics · 19/11/2020

  8. ... / (πd²/4) = 4*1100cm³/s / π(0.50cm)² = 5602 cm /s = 560 m/s v = 4*1100cm³/s / π(2.0cm)² = 350 cm /s = 35 m/s...

    1 Answers · Science & Mathematics · 20/10/2020

  9. ...283k = 1.01e5Pa * d³ / 293K solves to d = 0.89 cm Hope this helps!

    1 Answers · Science & Mathematics · 19/10/2020

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