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2 years ago

The drawing shows a tire of radius R on a moving car

Physics

1 answer:

masha68 [24]2 years ago

7 0

Answer:

H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

Em₀ = K = 1/2 m v²

Final. In the sought height

$Em_{f}$ = U = mg h

Energy is conserved

Em₀ = $Em_{f}$

½ m v² = m g h

v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

Em₀ = K = ½ I w²

Final. Heights sought

Emf = U = m g R

Em₀ = $Em_{f}$

½ I w² = m g R

The moment of inertial of a cylinder is

I = $I_{cm}$ + ½ m R²

I= ½ $I_{cm}$ + ½ m R²

Linear and rotational speed are related

v = w / R

w = v / R

We replace

½ $I_{cm}$ w² + ½ m R² w² = m g R

moment of inertia of the center of mass

$I_{cm}$ = ½ m R²

½ ½ m R² (v²/R²) + ½ m v² = m gR

m v² ( ¼ + ½ ) = m g R

v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

2 g H = 4/3 g R

H / R = 2/3

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2 years ago

Suppose that the Hubble Space Telescope discovers a series of planets with the following characteristics moving around a star th

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2 years ago

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Explanation:

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2 years ago

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2 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

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Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

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2 years ago