The power dissipated is simply V^2/R
where V = 120 volts RMS
and R = 60 Ω
Use the kinematics equation:
d = vt + 1/2at^2.
In this problem,
v = 15 m/s.
a = 3 m/s^2,
t = 10 s,
So:
d = 15(10) + 1/2*(3)*(10^2) = 300 meters
Answer:
Fm is the net force down on the metal in air
Fm / 2 is the net force down on the metal in liquid
Fl is the buoyant force on the metal due to liquid
Fm - Fl = Fm / 2 equating upward and downward forces
Fm / 2 = Fl
The specific gravity of the metal is twice that of the liquid
Note: F = M g = ρ V g since ρ = M / V
Answer:
The torque is approximately 3.82 N·m
Explanation:
The relationship between work done, 'W', and the applied torque, 'τ' can be presented as follows;
W = τ × Δθ
Where;
W = The work done by the torque
τ = The magnitude of the torque
Δθ = The angle through which the object is turned
The parameters for 'W' and 'Δθ' are;
W = 3j
Δθ = 1/4 × π
From W = τ × Δθ, we have;
τ = W/Δθ
∴ τ = 3j/(1/4 × π) ≈ 3.81971863421 N·m
The torque, τ ≈ 3.82 N·m