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2 years ago

How to determine the width/thickness of a coin using a screw gauge?

1 answer:

5 0

We can calculate the width/thickness of a coin using a screw gauge by placing it in between the teeth of the gauge.

<h3>How we can calculate the thickness using screw gauge?</h3>

We can determine the width/thickness of a coin by using a screw gauge because screw gauge is an instrument that measures thickness of an object that are very thin. First we can placed the coin in between the teeth of the gauge and then move the gauge until the coin can be held tightly. After that note the reading on the scale.

So we can conclude that we can calculate the width/thickness of a coin using a screw gauge by placing it in between the teeth of the gauge.

Learn more about thickness here: brainly.com/question/4937019

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Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially

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Answer:

Final speed is 900.06 m/s at $0.2215^{\circ}$

Solution:

As per the question:

Mass of the first asteroid, m = $15\times 10^{3}\kg$

Mass of the second asteroid, m' = $20\times 10^{3}\kg$

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, $\theta = 20^{\circ}$

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

$\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}$

Thus applying vector addition and momentum conservation, the final velocity is given by:

$(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}$ (1)

Now,

$(m +m')v_{final} = (35\times 10^{3})v_{final}$

$(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}$

$(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}$

$2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}$

Now, substituting the suitable values in eqn (1), we get:

$v_{final} = 900.06\ m/s$

Now, the direction for the two vectors is given by:

$\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}$

$\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}$

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How to determine the width/thickness of a coin using a screw gauge?​

How to determine the width/thickness of a coin using a screw gauge?