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3 years ago

14

A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r

esistance) the ball will land on the ground with a speed of
A. 0 m/s. B. 20 m/s. C. 30 m/s. D. 40 m/s. E. There is not enough information to say.

2 answers:

ICE Princess25 [194]3 years ago

6 0

B is the answer please mark brainliest

uysha [10]3 years ago

3 0

The answer is not B.

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Neon has 10 electrons; two in the inner level and 8 in its outermost level. Will neon form bonds with other atoms?

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Answer:

no

Explanation:

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4 years ago

How is Newton’s 3rd law in baseball?

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3 years ago

When wile coyote is launched his velocity is 20 m/s. As a slightly underweight 10kg coyote, how far would the 400 N/m spring hav

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Answer:

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3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of 0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = <em>0.27</em>

7 0

3 years ago