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3 years ago

Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.

Physics

1 answer:

tamaranim1 [39]3 years ago

4 0

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.

applying motion equation

(assume g = 10 m/s²)

$s=ut+\frac{1}{2}*gt^{2} \\= 0+\frac{1}{2}*10*1.081^{2}\\= 5.843 m$

Arrow misses the target by 5.843m ig the arrow us split horizontally

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The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr

stealth61 [152]

Answer:

$\frac{R_2}{R_1}=\frac{A_1}{A_2}\\\frac{R_4}{R_3}=\frac{A_3}{A_4}$

Explanation:

The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

$R=\frac{\rho L}{A}$

$\rho$ is the material's resistance, L is the legth and A is the cross-sectional area.

For the first and second coils, we have:

$R_1=\frac{\rho L}{A_1}\\R_2=\frac{\rho L}{A_2}\\\rho L=R_1A_1\\\rho L=R_2A_2\\R_1A_1=R_2A_2\\\frac{R_2}{R_1}=\frac{A_1}{A_2}$

For the third and fourth coils, we have:

$R_3=\frac{\rho L'}{A_3}\\R_4=\frac{\rho L'}{A_4}\\\rho L'=R_3A_3\\\rho L'=R_4A_4\\R_3A_3=R_4A_4\\\frac{R_4}{R_3}=\frac{A_3}{A_4}$

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3 years ago

What kind of wave is being generated?

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Generated for what there no waves shown

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4 years ago

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what is the amplitude and wavelength of the wave shown below? A. Amplitude: 15 cm; wavelength: 150 cm B. Amplitude: 30 cm; wavel

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A loudspeaker located in air generates sound waves of frequency 1,000 Hz. Some of these sound waves enter a pool of water, where

prisoha [69]

Answer:

1.4 m

Explanation:

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f = Frequency of sound = 1000 Hz

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The wavelength of the sound waves in water is 1.4 m

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3 years ago

What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency

Anvisha [2.4K]

Answer:

$1.04\times 10^{-7}$ T

Explanation:

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3 years ago