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AnnZ [28]
3 years ago
15

A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm

Physics
1 answer:
Effectus [21]3 years ago
8 0

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm

So, the thickness of the film is equal to 3.1 cm.

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two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire
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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

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In which medium does sound travel the fastest?
TEA [102]

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4 years ago
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What is the matching nitrogen base sequence for the gene below?
grigory [225]

Answer:

A

Explanation:

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6 0
3 years ago
What charge appears on each plate of a 10.0 μF parallel plate capacitor, when it is charged to 110 V?
elena-14-01-66 [18.8K]

Answer:

charge, q = ± 1.1 mC

Given:

Capacitance, C = 10.0\micro F = 10.0\times 10^{- 6} F

Voltage, V = 110 V

Solution:

The charge on the capacitor plates can be calculated by using the definition of capacitance as :

q ∝ V

where

q = charge

V = potential difference or Voltage

Therefore,

q = CV

Now, charge, q :

q = 10.0\times 10^{- 6}\times 110 = 1100\micro C = 1.1 mC

Therefore, the charge on the positive plate is:

q = + 1.1 mC

the charge on the negative plate is:

q = - 1.1 mC

8 0
3 years ago
What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w
Delvig [45]
(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by
f_1 =  \frac{1}{2L} \sqrt{ \frac{T}{m/L} }
where
L is the wire length
T is the tension
m is the wire mass

In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is
T= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }= \frac{1}{2 \cdot 10.9 m} \sqrt{ \frac{253 N}{0.0558 kg/10.9 m} }=    10.2 Hz

b) The frequency of the nth-harmonic for a standing wave in a wire is given by
f_n = n f_1
where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:
f_2 = 2 f_1 = 2 \cdot 10.2 Hz=20.4 Hz

c) Similarly, the third lowest frequency (third harmonic) is given by
f_3 = 3 f_1 = 3 \cdot 10.2 Hz = 30.6 Hz

8 0
3 years ago
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