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faltersainse [42]
3 years ago
8

Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range

Y: attractive and repulsive Z: very small range X: attractive only Y: very small range Z: attractive and repulsive X: attractive and repulsive Y: infinite range Z: attractive only X: very small range Y: attractive only Z: infinite range
Physics
1 answer:
Maslowich3 years ago
8 0

Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

X: Always attractive  or attractive only

Y: Very small range

Z:  Repulsive and attractive

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An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
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Answer:

1) A

2) C

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4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

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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}

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

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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}

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

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

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

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

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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

F = (3v)^{2}

F = 9v^{2}

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

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

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F = v^{2}

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F = (2v)^{2}

F = 4v^{2}

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

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

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F = 1/r

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F = 1/2 * 1/r

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

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