Time Power: A Proven System for Getting More Done in Less Time Than You Ever Thought Possible

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P = T A ω A = T B ω B , {\displaystyle P=T_{\text{A}}\omega _{\text{A}}=T_{\text{B}}\omega _{\text{B}},} The user is able to perform/cast a rare variety of magic spells and feats that manipulates time itself to a degree. P ( t ) = p Q , {\displaystyle P(t)=pQ,} where p is pressure in pascals or N/m 2, and Q is volumetric flow rate in m 3/s in SI units.

These relations are important because they define the maximum performance of a device in terms of velocity ratios determined by its physical dimensions. See for example gear ratios.If a mechanical system has no losses, then the input power must equal the output power. This provides a simple formula for the mechanical advantage of the system. Said projectile disrupts the local molecules by creating desynchronization, like a stutter, on a small scale. As time progresses, Chronon active individuals gain control over their powers, increasing the strength of existing abilities or adding new abilities to their repertoire. [1] Manifestation [ ] M A = F B F A = v A v B . {\displaystyle \mathrm {MA} ={\frac {F_{\text{B}}}{F_{\text{A}}}}={\frac {v_{\text{A}}}{v_{\text{B}}}}.} P = d W d t = d d t ∫ Δ t F ⋅ v d t = F ⋅ v . {\displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \,dt=\mathbf {F} \cdot \mathbf {v} .} Hence the formula is valid for any general situation. where ω is angular frequency, measured in radians per second. The ⋅ {\displaystyle \cdot } represents scalar product.

Main article: Electric power Ansel Adams photograph of electrical wires of the Boulder Dam Power Units, 1941–1942 W C = ∫ C F ⋅ v d t = ∫ C F ⋅ d x , {\displaystyle W_{C}=\int _{C}\mathbf {F} \cdot \mathbf {v} \,dt=\int _{C}\mathbf {F} \cdot d\mathbf {x} ,} V ( t ) {\displaystyle V(t)} is the potential difference (or voltage drop) across the component, measured in volts, and

The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second. Other common and traditional measures are horsepower (hp), comparing to the power of a horse; one mechanical horsepower equals about 745.7 watts. Other units of power include ergs per second (erg/s), foot-pounds per minute, dBm, a logarithmic measure relative to a reference of 1 milliwatt, calories per hour, BTU per hour (BTU/h), and tons of refrigeration.

P ( t ) = d W d t = F ⋅ v = − d U d t . {\displaystyle P(t)={\frac {dW}{dt}}=\mathbf {F} \cdot \mathbf {v} =-{\frac {dU}{dt}}.} In physics, power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. [1] [2] [3] Power is a scalar quantity. Chronon active individuals can manipulate the Chronon energy around them, depleting it from any given area and can use it to create shields, or "warp" across long distances in the blink of an eye. Chronon active users can create offensive projectiles that disrupt time, where it would otherwise flow normally. [1] p u l s e = ∫ 0 T p ( t ) d t {\displaystyle \varepsilon _{\mathrm {pulse} }=\int _{0}The ability to manipulate time is gained through exposure to above-average (and often dangerous) levels of Chronon energy. The resulting exposure creates what Monarch Solutions' Chronon department refers to as a " Chronon active" individual. [1] P = lim Δ t → 0 P a v g = lim Δ t → 0 Δ W Δ t = d W d t . {\displaystyle P=\lim _{\Delta t\to 0}P_{\mathrm {avg} }=\lim _{\Delta t\to 0}{\frac {\Delta W}{\Delta t}}={\frac {dW}{dt}}.}

Without the knowledge or proper skill, the intended use of such magic can create unpredictable or potentially dangerous results. Peak power and duty cycle [ edit ] In a train of identical pulses, the instantaneous power is a periodic function of time. The ratio of the pulse duration to the period is equal to the ratio of the average power to the peak power. It is also called the duty cycle (see text for definitions).P = d W d t = d d t ( F ⋅ x ) = F ⋅ d x d t = F ⋅ v . {\displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\left(\mathbf {F} \cdot \mathbf {x} \right)=\mathbf {F} \cdot {\frac {d\mathbf {x} }{dt}}=\mathbf {F} \cdot \mathbf {v} .} Mechanical power [ edit ] One metric horsepower is needed to lift 75 kilograms by 1 metre in 1 second. M A = T B T A = ω A ω B . {\displaystyle \mathrm {MA} ={\frac {T_{\text{B}}}{T_{\text{A}}}}={\frac {\omega _{\text{A}}}{\omega _{\text{B}}}}.}