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OverLord2011 [107]
3 years ago
14

A 2kg blob of putty moving at 4 m/s slams into a 6kg blob of putty at rest what is the speed of the to stick together blobs imme

diately after colliding
Physics
1 answer:
andrew11 [14]3 years ago
3 0
<h2>Answer</h2>

1m/s

<h2>Explanation</h2>

Given that:

<em>Mass of first blob = 2kg = m1</em>

<em>Velocity of blob = 4m/s = v1</em>

<em>Mass of second blob = 6kg = m2</em>

<em>Velocity of blob = 0m/s = v2</em>

<em />

To find:

<em>Final velocity = Vf</em>

<em />

<em>This question is of inelastic collision which is any collision between objects in which some energy is lost.</em>

<em />

<h3>Formula to be use:</h3><h2>(m1*v1) + (m2*V2) = Vf(m1 + m2)</h2>

(2*4) + (6*0) = Vf(2+6)

8 + 0 = Vf(8)

8 = Vf(8)

Vf = 1 m/s

So the speed of two blobs immediately after colliding = 1 m/s

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Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re
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Answer:

K_{system} = \frac{k}{10}

Explanation:

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\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})

Here, n = 10

Spring constant of each spring = k

Thus,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})

\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})

\frac{1}{K_{system}} = \frac{10}{k}

K_{system} = \frac{k}{10}

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An electron is accelerated through 2.40×10³V from rest and then enters a uniform 1.70-T magnetic field. What are(b) the minimum
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The maximum value of Force is 7.88x10^{-12}

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we need to find

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b) The minimum value of magnetic force

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