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aniked [119]
4 years ago
12

1. A baseball is thrown vertically at 16.7 m/s. What is the maimum height of the baseball?

Physics
1 answer:
Anvisha [2.4K]4 years ago
7 0

Answer:

14.2 m

Explanation:

Using conservation of energy:

PE at top = KE at bottom

mgh = ½ mv²

h = v² / (2g)

h = (16.7 m/s)² / (2 × 9.8 m/s²)

h = 14.2 m

Using kinematics:

Given:

v₀ = 16.7 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 14.2 m

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A parallel-plate air capacitor is connected to a constant-voltage battery. If the separation between the capacitor plates is dou
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Drop to half of the previous value

Explanation:

Energy stored in capacitor is inversly propotional to the distance between the plates.

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The Tampa Bay area in Florida received insufficient precipitation over a long period of time. This resulted in a drought in the
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Read 2 more answers
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =
Sloan [31]

Answer:

The current is  I  = 8.9 *10^{-5} \  A

Explanation:

From the question we are told that

     The  radius is r =  3.17 \  mm  =  3.17 *10^{-3} \ m

      The current density is  J =  c\cdot r^2  =  9.00*10^{6}  \ A/m^4 \cdot r^2

      The distance we are considering is  r =  0.5 R  =  0.001585

Generally current density is mathematically represented as

          J  =  \frac{I}{A }

Where A is the cross-sectional area represented as

         A  =  \pi r^2

=>      J  =  \frac{I}{\pi r^2  }

=>    I  =  J  *  (\pi r^2 )

Now the change in current per unit length is mathematically evaluated as

        dI  =  2 J  *  \pi r  dr

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         I  = 2\pi  \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr

         I  = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr

        I  = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ]  | \left    0.001585} \atop 0}} \right.

        I  = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]

substituting values

        I  = 2 *  3.142  *  9.00 *10^6 *   [ \frac{0.001585^4}{4} ]

        I  = 8.9 *10^{-5} \  A

5 0
4 years ago
Question 2 (5 points)
d1i1m1o1n [39]

Answer:

Frequency of sound wave = 198.83 hertz (Approx.)

Explanation:

Given:

Velocity of sound wave in air = 340 m/s

Wavelength = 1.71 meter

Find:

Frequency of sound wave

Computation:

Frequency = Velocity / Wavelength

Frequency of sound wave = Velocity of sound wave in air / Wavelength

Frequency of sound wave = 340 / 1.71

Frequency of sound wave = 198.8304

Frequency of sound wave = 198.83 hertz (Approx.)

6 0
3 years ago
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