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Svetradugi [14.3K]
2 years ago
6

A 3000-N force gives an object an acceleration of 15 m/s2. The mass of the object is

Physics
1 answer:
7nadin3 [17]2 years ago
5 0
<h3>Answer:</h3>

200 kg

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Physics</u>

<u>Newton's Law of Motions </u>

Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion

Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)

Newton's 3rd Law of Motion: For every action, there is an equal and opposite reaction<u> </u>

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] F = 3000 N

[Given] a = 15 m/s²

[Solve] m = <em>x</em> kg

<u>Step 2: Solve for </u><em><u>m</u></em>

  1. Substitute in variables [Newton's Second Law of Motion]:                          3000 N = m(15 m/s²)
  2. [Mass] [Division Property of Equality] Isolate <em>m</em> [Cancel out units]:            200 kg = m
  3. [Mass] Rewrite:                                                                                               m = 200 kg
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7 0
3 years ago
Starting at t = 0 s , a horizontal net force F⃗ =( 0.285 N/s )ti^+(-0.460 N/s2 )t2j^ is applied to a box that has an initial mom
Irina-Kira [14]

Answer:

Explanation:

We know that Impulse = force x time

impulse = change in momentum

change in momentum = force x time

Force F = .285 t -.46t²

Since force is variable

change in momentum = ∫ F dt  where F is force

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

When t = 1.9

change in momentum = .285 x 1.9² /2 i  -  .46 x 1.9³ / 3 j

= .514i - 1.05 j

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= - 3.1 i + 3.9j +.514i - 1.05j

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7 0
3 years ago
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Leno4ka [110]

Explanation:

Given that,

Mass of the car, m₁ = 1250 kg

Initial speed of the car, u₁ = 7.39 m/s

Mass of the truck, m₂ = 5380 kg

It is stationary, u₂ = 0

Final speed of the truck, v₂ = 2.3 m/s

Let v₁ is the final velocity of the car. Using the conservation of momentum as :

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So, the final velocity of the car is 2.5 m/s but in opposite direction. Hence, this is the required solution.

3 0
3 years ago
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