<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
Answer:
Yes Newton's laws apply to small objects
EX: Newton s first law
when body at rest always want to be at rest
or body at motion always want to be at motion
unles an external force acts upon it
for example a eraser on the table will be at rest
if so e apply some force then it comes motion
so, Newton s law apply to small object s
Answer:
≅279.02 N, 36° 15' from North to East
Explanation:
You can the theory of triangle of forces to find the answer.
Two forces are acting perpendicular to each other so use Pythagoras theorem to find the magnitude of the resultant force(x).
x² = 225² + 165²
x = 279.02 N
To find the direction use the same force triangle,
Workings are in the attachment,
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀