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2 years ago

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The greatest speed recorded by a baseball thrown by a pitcher was 162.3 km / h, obtained by Nolan Ryan in 1974. If the ball leav

es the pitcher's hand with a horizontal speed of this magnitude, how much will the ball have fallen? by the time you have traveled 20 m horizontally?

1 answer:

Pepsi [2]2 years ago

3 0

Answer:

0.96 m

Explanation:

First, convert km/h to m/s.

162.3 km/h × (1000 m/km) × (1 hr / 3600 s) = 45.08 m/s

Now find the time it takes to move 20 m horizontally.

Δx = v₀ t + ½ at²

20 m = (45.08 m/s) t + ½ (0 m/s²) t²

t = 0.4436 s

Finally, find how far the ball falls in that time.

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.4436 s) + ½ (-9.8 m/s²) (0.4436 s)²

Δy = -0.96 m

The ball will have fallen 0.96 meters.

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Explanation:

We know that charge on electron

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Now by putting the value

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

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Explanation:

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

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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:

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

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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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