Steam enters a cylinder—- A
Answer: 1896.55J/kg°C
Explanation:
The quantity of Heat Energy (Q) required to heat a material depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = 1320 joules
Mass of material = 5.61kg
C = ? (let unknown value be Z)
Φ = 0.124°C
Then, Q = MCΦ
1320J = 5.61kg x Z x 0.124°C
1320J = 0.696kg°C x Z
Z = (1320J / 0.696kg°C)
Z = 1896.55 J/kg°C
Thus, the specific heat of the material is 1896.55J/kg°C
Use the Pythagoras for the magnitude and the tan^-1 x = -1 for the angle
displacement = 4^2 + 4^2 = 32 = 4 sqrt(2) = 5.65 km
angle is 135 degrees.