electric flat cart solution 1-500 t

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(b) Gaussian surface for calculating the electric field. Solution: The electric field is non-vanishing only in the region ar<

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500 2− f 500 df = 1 500 f2 2 500 0 +2f 1000 500 − 1 500 f2 2 1000 500 = 1 500 × 1 2 1 500 × 1 2 (106 −25×10) = 250+1000−750 = 500 W m−2. The total power density of the radiation is 500 W/m2. b. What is the value of the bandgap of the photodiode mate-rial that results in the maximum theoretical efficiency of the photodiode exposed

The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. T TI 1 22 2 0 1 2 = ... SOLUTION Angular speeds. wp ww wp ...

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Electric cars or all-electric cars are a type of electric vehicle (EV) that has a rechargeable battery pack onboard that can be charged from the electric grid, and the electricity stored on the vehicle is the only source that drives the wheels for propulsion. The term "electric car" generally refers to highway-capable automobiles, but there are

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T= 2ˇ! = 2ˇ 8 = ˇ 4 seconds: The frequency is the number of cycles that the mass makes in 1 second. For the example above it is f= 1 T =! 2ˇ = cycles per second !(in x00= !2x) is called the circular frequency. For a mass spring system,!2 = k m The circular frequency for the example above is,!= r k m = cycles per second 7

Solution: P = 4.5hp = (4.50)(746 watts) = 3360 watts. P = W/t ; W = Pt ; W = 3360J/s (36,000s) = 1.21 x 10 8 J. Efficiency: When a device receives power from a source, it does not deliver all of it in the intended form and converts a portion of it to other undesired forms.

gets deflected by 1.0 cm in the y direction what is v 0 (in m/s)? Answer: 1.05 × 107 Solution: The deflection by a constant acceleration is 2 y=1at2, where y = 0.05 m, the acceleration a=Ee/m e and the time t=Δx/v 0. Solving yields v 0 = 1.05 × 10 7 m/s. 5. A conducting sphere with charge +10 nC is placed at the center of two

Calculate the electric potential associated with the proton’s electric field at this distance. Solution The potential of the proton, at the position of the electron (both of which may be regarded as point-charge atomic constituents) is (Equation 25-4) V = ke=a 0, where a 0 is the Bohr radius. Numerically, V = (9 ×10 9 N ⋅ m2/C2) × (1.6 × 10