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3 years ago

If a cannonball were launched with a speed of 37m/s at a 60° angle from the height of Om, what would be the cannonball's range?

Physics

1 answer:

JulijaS [17]3 years ago

8 0

Answer:

Range = 120.98 m, (or rounded to the nearest whole number =121 meters)

Explanation:

We can use the formula for range in projectile motion at ground level:

$Range = \frac{v^2\,\,sin(2\theta)}{g} \\Range= \frac{37^2\,\,sin(120)}{9.8} \\Range\approx 120.98\,\,m$

Then the cannon's range would be approximately 120.98 m, which may be rounded to the nearest whole number as 121 meters

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Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

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Think about a typical school day. In the space provided below, describe how each of the different forms of energy we have learne

Nezavi [6.7K]

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AleksAgata [21]

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The following is an example of how humans can increase biodiversity

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