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1 year ago

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An amusement park ride consists of airplane-shaped cars attached to steel rods. Each rod has a length of 15m and a cross-section

al area 8.00cm^2. Take Young's modulus for steel to be Y= 2.0 x 10^11 Pa.
How much is the rod stretched (change in length of) when the ride is at rest? Assume that each car with two people seated in it has a total weight of 1900 N.

1 answer:

Oliga [24]1 year ago

7 0

Explanation:

Given that,

Length of each rod, L = 15 m

Area of cross section of the rod, $A=8\ cm^2=0.0008\ m^2$

Young's modulus for steel, $Y=2\times 10^{11}\ Pa$

Total weight of two people, F = 1900 N

We need to find the stretching in the rod when the ride is at rest. The formula of Young's modulus of a material is given by :

$Y=\dfrac{FL}{A\Delta L}$

Here, $\Delta L$ is the stretching in the rod

$\Delta L=\dfrac{FL}{YA}\\\\\Delta L=\dfrac{1900\times 15}{2\times 10^{11}\times 0.0008}\\\\\Delta L=1.78\times 10^{-4}\ m$

So, the rod will stretched to $1.78\times 10^{-4}\ m$

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Explanation:

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2 years ago

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We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

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Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

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1 year ago

The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan

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Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u> proportional to the current and <u><em>inversely</em></u> proportional to the distance from the wire. If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

Magnetic field around a long current carrying wire is given by

$B=\frac{\mu _o I}{2\pi r}$

where B= magnetic field

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I= current in the long wire and

r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u> proportional to the current and <u><em>inversely</em></u> proportional to the distance from the wire.

Now if I'=3I and r'=2r then magnetic field B' is given by

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1 year ago

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Answer:

a) 1.2*10^-7 m

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(b)

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Answer:

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Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

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Substituting, we find:

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1 year ago