A pendulum that moves through its equilibrium position once every 1.000 s is sometimes called a "seconds pendulum."

1 answer:

3 0

<span>I would solve this problem by starting from the equation for a pendulum's period:

T ≈ 2π√( L / g )

Rearranging for g, and substituting the length in Cambridge:

g ≈ ( 4π² * L ) / T²
. . = ( 4π * 0.9942 m ) / ( 1.000 s + 1.000 s )²
. . = 9.812 m/s²

And for Tokyo:

g ≈ ( 4π² * L ) / T²
. . = ( 4π * 0.9927 m ) / ( 1.000 s + 1.000 s )²
. . = 9.798 m/s²</span>

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A pendulum with a length of 1 meter is released from an initial angle of 15.0° after 1000s its amplitude has been reduced by fri

Setler [38]

Answer:

0.366×10^{-3} / s

Explanation:

θ = θmax e^{-bt/2m}

Given that

θ = 5.50°

θmax = 15.0°

So that we have

ln (θ / θmax) = -bt /2m

= - ln(5.50°/ 15.0°) / 1000s = b /2m

= b / 2m = 0.366×10^{-3} / s

3 0

3 years ago

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$