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11111nata11111 [884]
2 years ago
7

A cannonball is fired from a cliff that is 50 meters above the ground. The cannonball is fired horizontally with a speed of 120

meters per second.
Calculate the horizontal distance that the cannonball will travel.
Physics
1 answer:
tankabanditka [31]2 years ago
4 0

The horizontal distance that the cannonball will travel is 383.33m

In order to find the horizontal distance, we will use the formula for calculating the range expressed as:

R = u\sqrt{\frac{2H}{g} } where:

u is the velocity

H is the height

g is the acceleration due to gravity

R = 120\sqrt{\frac{2(50)}{9.8} }\\R = 120\sqrt{\frac{100}{9.8} } \\R=120\times 3.1943\\R=383.33m

Hence the horizontal distance that the cannonball will travel is 383.33m

Learn more here: brainly.com/question/19028766

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1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = 10ms^{-2}, F = ma = 15*10 = 150N

2) m = 3kg, v = 4ms^{-1}, r = 4m, F = \frac{mv^{2} }{r}

\frac{3*4^{2} }{4} = 12N

3) a = 10ms^{-2}, r = 10m, v=?

F = \frac{mv^{2} }{r} and F = ma

equating the two equations and cancelling a, we have:

\frac{v^{2} }{r} = a

making v the subject of formula, we have:

v = \sqrt{ar}

= \sqrt{100}

= 10ms^{-1}

4) r = 10m, v = 5ms^{-1}, a = ?

F = \frac{mv^{2} }{r}

F = ma

equating the above equations and making a subject of formula, we get:

a = \frac{v^{2} }{r}

a = 25/10 = 2.5ms^{-2}

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6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is tripled

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We can see that the force will be 9X greater than it was.

9) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is doubled

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F = 4v^{2}

We can see that the force will be 4X greater than it was.

10) F = \frac{mv^{2} }{r}

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

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3 years ago
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