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maks197457 [2]
3 years ago
6

You are given a long length of string and an oscillator that can shake one end of the string at any desired freqeuency

Physics
1 answer:
Anastasy [175]3 years ago
7 0
Based on this given information, the oscillator can perform a simulated oscillation on the given string. However, this is dependent on the speed and length of the oscillator and string.Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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There are no risks when taking medication to treat anxiety.
lidiya [134]

Answer:

False

Explanation:

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A mechanical device requires 500j of work to do 200j of work in lifting a box. What is the efficiency of the device?
olga55 [171]
40% the formula is work output/work input*100 so
200/500=.4*100=40
5 0
3 years ago
A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o
Ierofanga [76]

Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: <span>

<span>Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is: 

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx. 

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have: 

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx. 

<span>Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)</span>

<span>W = ∫ 1250π x dx  
<span>W = 1250π ∫ x dx
W = 625π * (x2 – x1)</span></span></span></span>

<span>x2 = 14 ft  + 15 ft = 29 ft</span>

x1 = 14 ft + 1 ft = 15 ft

<span>
W = 625π * (29^2 - 15^2) 
<span>W = 385,000π ft-lbs = 1,209,513.17 ft-lbs</span></span>

3 0
4 years ago
A 9.12 microFarad capacitor is measured to have a reactance of 268.0 Ohm. At what frequency (in Hz) is it being driven?
Katen [24]

Answer:

65.14 Hz

Explanation:

We have given the capacitance C=9.12\mu F=9.12\times 106{-6}F

And the capacitive reluctance = 268 ohm

The capacitive reluctance of a capacitive circuit is given by

X_C=\frac{1}{\omega C}

X_C=\frac{1}{2\pi f C}

268=\frac{1}{2\pi f \times 9.12\times 10^{-6}}

f=\frac{1}{2\times 3.14\times 268\times 9.12\times 10^{-6}}=65.14Hz

3 0
4 years ago
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