The slope of the line is
(change in ' y ' between the ends) / (change in ' x ' between the ends)
Slope = (630g - 0) / (70 cm^3 - 0)
Slope = (630 / 70) g/cm^3
<em>Slope = 9.0 g/cm^3</em>
Given: Mass m = 44 Kg; Velocity v = 10 m/s
Required: Kinetic energy K.E = ?
Formula: K.E = 1/2 mv²
K.E 1/2 (44 Kg)(10 m/s)²
K.E = 2,200 Kg.m²/s²
K.E = 2,200 J Answer is A
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Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
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