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3 years ago

A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the

Physics

1 answer:

Umnica [9.8K]3 years ago

6 0

Answer:

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

$H_{max}=\frac{V^2\sin ^2 \theta}{2g}$

g = 9.8m/s2

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

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<h3>Newton's law of universal gravitation</h3>

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The mathematical interpretation of the above law is

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Where:

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From the above, When you attract every object in the universe with a force that is proportional to the mass of the objects and to the distance between them, we are obeying Newton's law of universal gravitation.

Learn more about Newton's law of universal gravitation here: brainly.com/question/9373839

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2 years ago

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Zinaida [17]

Answer:

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Here k is the Coulomb constant. In this case, we have $q_1=-e$ , $q_2=e$ and $d=4.09*10^-10m$ . Replacing the values:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N$

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

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svetoff [14.1K]

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