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2 years ago

What is the magnitude of velocity for a 2,000 kg car possessing 3,000 kg(*)m/s of momentum?

Physics

1 answer:

Trava [24]2 years ago

5 0

The magnitude of velocity for this car is equal to 1.5 m/s.

<u>Given the following data:</u>

Momentum of car = 3,000 kgm/s.

Mass of car = 2,000 kg.

To calculate the magnitude of velocity for this car:

<h3>What is momentum?</h3>

In Science, momentum simply means a multiplication of the mass of an object and its velocity.

Mathematically, momentum is giving by the formula;

$Momentum = mass \times velocity$

Making velocity the subject of formula, we have:

$Velocity=\frac{Momentum}{Mass}$

Substituting the given parameters into the formula, we have;

$Velocity=\frac{3000}{2000}$

Velocity = 1.5 m/s.

Read more on momentum here: brainly.com/question/15517471

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If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to

-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

$\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm$

Moment of inertia from neutral axis will be given by $\frac {bd^{3}}{12}+ ay^{2}$

Therefore, moment of inertia will be

$\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}$

Bending stress at top= $\frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa$

Bending stress at bottom= $\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03$ Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

8 0

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Answer:

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