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sp2606 [1]
3 years ago
9

What is the frequency of a pendulum of length 1.50 m at a location where 1 point the acceleration due to gravity is 9.79 m/s^2?

(T=2TT x square root of the length/gravity) and (f = 1/T).*
Physics
1 answer:
ivann1987 [24]3 years ago
4 0

Answer:

Explanation:The simple pendulum calculator finds the period and frequency of a ... Acceleration of gravity (g) ... Pendulum length (L) ... First of all, a simple pendulum is defined to be a point mass or bob (taking ... For example, it can be equal to 2 m. ... Find the frequency as the reciprocal of the period: f = 1/T = 0.352 Hz

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17. What is the speed of a race horse that runs<br> 1500 m in 125 s?
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3 years ago
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A parallel plate capacitor filled with air with plate area 2-cm2 and plate separation of 0.5 mm is connected to a 12 V batter an
pashok25 [27]

Answer:

a) The charge of the capacitor is 4.25x10⁻¹¹C

b) The charge of the capacitor is 4.25x10⁻¹¹C because the battery is disconnected.

c) The potential difference across the plates is 18 V

d) The work is 7.64x10⁻¹⁰J

Explanation:

The capacitance of the capacitor is equal to:

C=\frac{e_{0}A }{d}

A = 2 cm² = 0.0002 m²

d = 0.5 mm = 0.0005 m

Replacing:

C=\frac{8.85x10^{-12}*0.0002 }{0.0005} =3.54x10^{-12} F = 3.54pF

a) The charge of the capacitor is equal to:

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b) The charge is the same because the battery is disconnected (Q = 4.25x10⁻¹¹C)

c) If distance is increased, we have:

C=\frac{8.85x10^{-12}*0.0002 }{0.00075} =2.36x10^{-12} F=2.36pF

The potential is:

V=\frac{Q}{C} =\frac{42.48}{2.36} =18V

d) The work done is equal to:

W=VQ=18*42.48=764.6pJ=7.64x10^{-10} J

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3 years ago
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a 10-N force is exerted on a box, moving it 20 m in the same direction. What is the magnitude of work done on the box?
-Dominant- [34]
F = 10 N

d = 20 m

θ = 0°

W = F (dot product) D = F * D * cos(angle between them)

W = FDcosθ

W = 10 * 20 * cos0 = 200 J
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3 years ago
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