We read your photo as:

Determine the dimensional equations for power, energy density, voltage, and permittivity.

Physics

ScanSolve AI Answer

Step-by-step solution

4 steps

Here's the solution to the problems:

b) (i)

  1. Define power in terms of work and time.

    Power (PP) is the rate at which work (WW) is done.

    P=WtP = \frac{W}{t}
  2. Express work in terms of force and distance, and force in terms of mass and acceleration.

    Work (WW) = Force (FF) ×\times Distance (LL)
    Force (FF) = Mass (MM) ×\times Acceleration (aa)
    Acceleration (aa) = DistanceTime2=LT−2\frac{\text{Distance}}{\text{Time}^2} = L T^{-2}

  3. Substitute the fundamental quantities (Mass MM, Length LL, Time TT) into the equations.

    F=M×(LT−2)=MLT−2F = M \times (L T^{-2}) = M L T^{-2} W=(MLT−2)×L=ML2T−2W = (M L T^{-2}) \times L = M L^2 T^{-2} P=ML2T−2T=ML2T−3P = \frac{M L^2 T^{-2}}{T} = M L^2 T^{-3}

    The dimensional equation of power is ML2T−3\boxed{M L^2 T^{-3}}.

    b) (ii)

  4. Determine the dimensions of each quantity in the LMTI system.

    Energy density (WdW_d) is energy per unit volume.
    Energy (EE) has dimensions ML2T−2M L^2 T^{-2}.
    Volume (VolVol) has dimensions L3L^3.

    [Wd]=[E][Vol]=ML2T−2L3=ML−1T−2[W_d] = \frac{[E]}{[Vol]} = \frac{M L^2 T^{-2}}{L^3} = M L^{-1} T^{-2}

    Distance (dd) has dimensions LL.

    [d]=L[d] = L

    Voltage (VV) is work per unit charge. Charge (QQ) = Current (II) ×\times Time (TT).

    [V]=[W][Q]=ML2T−2IT=ML2T−3I−1[V] = \frac{[W]}{[Q]} = \frac{M L^2 T^{-2}}{I T} = M L^2 T^{-3} I^{-1}

    Permittivity (ϵ\epsilon) can be derived from Coulomb's law $F

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp