<h3>
Answer:</h3>
Magnitude of Impulse: 30000 kg · m/s or 30000 N · s
Force on the Car: -6000 N
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality<u>
</u>
<u>Physics</u>
<u>Momentum</u>
- Momentum Equation: P = mv
- Impulse Equation: J = FΔt
- Law of Conservation of Momentum
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
Mass <em>m</em> = 1.50 × 10³ kg
Velocity <em>v</em> = 20 m/s east
Change in time Δt = 5.00 s
<u>Step 2: Find Magnitude</u>
- Substitute [Momentum]: P = (1.50 × 10³ kg)(20 m/s)
- Multiply: P = 30000 kg · m/s
<u>Step 3: Find Force</u>
<em>We use the Law of Conservation of Momentum to find our break force acting upon the car.</em>
- Substitute [Impulse]: 30000 kg · m/s = F(5.00 s)
- Rewrite: 30000 N · s = F(5.00 s)
- Divide 5 on both sides: 6000 N = F
- Rewrite: F = 6000 N
Since the car is deaccelerating, the break force would be towards the west direction (negative as east is our positive direction).
∴ F = -6000 N
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Answer:
Explanation:
1) ½(0.8)t² = 4t
0.4t² = 4t
0.4t = 4
t = 10 s
2) d = vt = 4(10) = 40 m
d = ½atⁿ = ½(0.8)10² = 40 m
3) v = at = 0.8(10) = 8 m/s
