"Which gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string?"

Answers

Answer 1

Answer:

the second derivative of y with respect to time gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string

Explanation:

This is because the transverse wave movement of particles take place in direction 90° to direction of movement of the wave (x) itself, so second derivative of y with respect to time (t)is what will be required

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An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.

Answers

The new length of aluminum rod is 17.435 cm.

The linear expansion coefficient is given as,

$\alpha=(L_(1)-L_(0))/(L_(0)(T_(1)-T_(0)))$

Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.

and linear expansion coefficient is $25*10^(-6)C^(-1)$

Substitute, $L_(0)=17.400cm,T_(1)=100,T_(0)=20,\alpha=25*10^(-6)C^(-1)$

$25*10^(-6)C^(-1) =(L_(1)-17.400)/(17.400(100-20))\n\n25*10^(-6)C^(-1) = (L_(1)-17.400)/(1392) \n\nL_(1)=[25*10^(-6)C^(-1) *1392}]+17.400\n\nL_(1)=17.435cm$

Hence, The new length of aluminum rod is 17.435 cm.

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Answer:

the new length is 17.435cm

Explanation:

the new length is 17.435cm

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A point charge that is exactly q =3 mu or micro CC is at the origin. In this problem, assume that the Coulomb constant k = 8.99 times 109 N m2/C2 exactly.(a) Find the potential V on the x axis at x = 3.00 m and at x = 3.01 m. In this part, enter your answers to exactly 6 signfificant figures.

Answers

Answer:

a) 8,990.00 V b) 8,960.13 V

Explanation:

a) The potential due to a point charge, can be found from the expression of Coulomb's Law, as follows:

$V = (k*q)/(r)$

where k = 8.99*10⁹ N*m²/C², q = 3.00*10⁻⁶ C, and r = 3.00 m.

Replacing by this values, we can find the potential V as follows:

$V = (8.99e9 N*m2/C2*3.00e-6C)/(3.00m) = 8,990.00 V$

b) Repeating the process for r = 3.01m:

$V = (8.99e9 N*m2/C2*3.00e-6C)/(3.01m) = 8,960.13 V$

Final answer:

The potential V at x=3.00 m and x=3.01 m from a point charge at the origin is 8.99 * 10^3 V and 8.97 * 10^3 V, respectively. This calculation is based on Coulomb's Law.

Explanation:

The potential V at a distance r from a point charge q is given by Coulomb's Law:

V = k*(q/r)

Here, k = 8.99 * 10^9 N*m^2/C^2 is the Coulomb constant, q = 3 µC is the charge, and r is the distance from the origin along the x-axis. For x = 3.00 m and x = 3.01 m, we can substitute these values into the equation to find V:

V1 = ((8.99 * 10^9 N*m^2/C^2)* (3 * 10^-6 C)) / 3.00 m = 8.99 * 10^3 V (to 6 significant figures)
V2 = ((8.99 * 10^9 N*m^2/C^2)* (3 * 10^-6 C)) / 3.01 m = 8.97 * 10^3 V (to 6 significant figures)

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If 3.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.59 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.

Answers

Answer:

0.158 A.

Explanation:

Mass of gold deposited = 3 x 10^-3 kg

= 3 g

Molar mass = 196 g/mol

Number of moles = 3/196

= 0.0153 mol.

Faraday's constant,

1 coloumb = 96500 C/mol

Quantity of charge, Q = 96500 * 0.0153

= 1477.04 C.

Remember,

Q = I * t

t = 2.59 hr

= 2.59 * 3600 s

= 9324 s

Current, I = 1477.04/9324

= 0.158 A.

Answer:

0.158A

Explanation:

Using Faraday's first law of electrolysis which states that the mass(m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of charge or electricity (Q) passed. i.e

m ∝ Q

m = Z Q

Where;

Z is the proportionality constant called electrochemical equivalent.

Faraday also observed that when 1 Faraday of electricity is equivalent to 96500C of charge.

Also,

Quantity of charge (Q), which is the product of current (I) passing through and the time taken (t) for the electrolysis, is given by;

Q = I x t; ----------------------(i)

With all of these in place, now let's go answer the question.

Since the gold ions carry one elementary unit of positive charge, now let's write the cathode-half reaction for gold (Au) as follows;

Au⁺ + e⁻ = Au ---------------------(ii)

From equation (ii) it can be deduced that when;

1 Faraday (96500C) of electricity is passed, 1 mole of Au forms ( = 197 grams of Au) [molar mass of Au = 197g]

Then, 3.00 x 10⁻³ kg (= 3 g of Au) will be formed by 3g x 96500C / 197g = 1469.5C

Therefore, the quantity of charge (Q) deposited is 1469.5C

Substitute this value (Q = 1469.5C) and time t = 2.59h (= 2.59 x 3600 s) into equation (i);

Q = I x t

1469.5 = I x 2.59 x 3600

1469.5 = I x 9324

Solve for I;

I = 1469.5 / 9324

I = 0.158A

Therefore, the current in the cell during that period is 0.158A

Note:

1 mole of gold atoms = 176g

i.e the molar mass of gold (Au) is 176g

A 13.0-Ω resistor, 13.5-mH inductor, and 50.0-µF capacitor are connected in series to a 55.0-V (rms) source having variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.

Answers

Answer:

E = 0.13 J

Explanation:

At resonance condition we have

$\omega = \sqrt{(1)/(LC)}$

$\omega = \sqrt{(1)/((13.5 * 10^(-3)){50* 10^(-6))}}$

$\omega = 1217.2 rad/s$

now if the frequency is double that of resonance condition then we have

$x_L = 2\omega L$

$x_L = 2(1217.2)(13.5* 10^(-3))$

$x_L = 32.86 ohm$

now we have

$x_c = (1)/(2(1217)(50* 10^(-6)))$

$x_c = 8.22 ohm$

now average power is given as

$P = i_(rms)V_(rms)* (R)/(z)$

$P = (55)/(√((32.86 - 8.22)^2 + 13^2)))(55)* (13)/(√((32.86 - 8.22)^2 + 13^2))$

$P_(avg) = 50.67 Watt$

Now time period is given as

$T = (2\pi)/(\omega)$

so total energy consumed is given as

$E_(avg) = 50.67((2\pi)/(2(1217.2)))$

$E = 0.13 J$

A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?

Answers

Answer:

$R\approx12.43 \,\, \Omega$

Explanation:

We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:

$V=I\,\,R\nR = (V)/(I) \nR=(87)/(7) \nR\approx12.43 \,\, \Omega$

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

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At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the tires rotating? Part (b) What is the centripetal acceleration at the edge of the tire in m/s^2?