You need to have the Mass and velocity
Answer:
Q=mc∆tita.
4780J= m × 2445J/kg/°C × 5.43°C
4780J= m× 13276.35J/kg
m= 0.36Kg
Answer:

General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Geometry</u>
- Area of a Rectangle: A = lw
<u>Algebra I</u>
- Exponential Property:

<u>Calculus</u>
Derivatives
Differentiating with respect to time
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Explanation:
<u>Step 1: Define</u>
Area is A = lw
2w = l
w = 300 m

<u>Step 2: Rewrite Equation</u>
- Substitute in <em>l</em>: A = (2w)w
- Multiply: A = 2w²
<u>Step 3: Differentiate</u>
<em>Differentiate the new area formula with respect to time.</em>
- Differentiate [Basic Power Rule]:

- Simplify:

<u>Step 4: Find Rate</u>
<em>Use defined variables</em>
- Substitute:

- Multiply:

- Multiply:

Answers: 1) 3 kg m²
2) 2.88 kg m²
Explanation: <u> </u><u>Question 1</u>
I = m(r)²+ M(r)²
I = 1.2 kg × (1 m )² +1.8 kg ×(1 m )²
∴ I = 3 kg m²
<u> </u><u>Question 2 </u>
ACCORDING TO THE DIAGRAM DRAWN FOR QUESTION 2
we have to decide where the center of gravity (G) lies and obviously it should lie somewhere near to the greater mass.<em> (which is 1.8 kg). S</em>ince we don't know the distance from center of gravity(G) to the mass (1.8 kg) we'll take it as 'x' and solve!!
<u>moments around 'G' </u>
F₁ d ₁ = F₂ d ₂
12 (2-X) = 18 (X)
24 -12 X =18 X
∴ X = 0.8 m
∴ ( 2 - x ) = 1.2 m
∴ Moment of inertia (I) going through the center of mass of two masses,
⇒ I = m (r)² +M (r)²
⇒ I = 1.2 × (1.2)² + 1.8 × (0.8)²
⇒ I = 1.2 × 1.44 + 1.8 × 0.64
⇒ I = 1.728 + 1.152
⇒ ∴ I = 2.88 kg m²
∴ THE QUESTION IS SOLVED !!!
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