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PHYC 1290 Lecture Notes - Lecture 20: Junkers J.I

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1 Jan 2019

School

Dalhousie University

Department

Physics & Atmospheric Science

Course

Professor

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Document Summary

,_- - - -+ -:__=~~ ~ - ~~ - - t l ! l , -- , ----+----"---. ,___ ____ _ ---r-~. --:----;- - - - - - - - - - - , -- - - -+ -__ (:);_\_t_ -=6:,b+. (!,-=c"--+- ~ __ h_tn-l_ -_ _,,__c _,o f---r_i \_e

Related textbook solutions

Physics: Principles with Applications

7th Edition, 2013

ISBN: 9780321625922

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Related Questions

The figure illustrates flow through a pipe with diameters of 1.0mm and 2.0 mm and with different elevations. Px is the pressure inthe pipe, and Vx is the speed of a non-viscous incompressible fluidat locations x = Q,R,S,T, or U.

Answer Choices: G-Greater Than, L-Less Than, E-Equal To

(a) VS is ... VQ

(b) VU is ... VR

(c) PT is ... PS

(d) PQ is ... PS

(e) VU is ... 2VT

(f) PS is ... PU.

calculate the angular momentum of the earth in its orbit around the sun?

m   5.97 10 kg 6 E R   6.38 10 m Orbital radius 11 r   1.50 10 m Period of rotation Prot = 24 h 86,400 s  Period of revolution Prev = 7 1 y 3.156 10 s   Identify: L I z z   Set Up: For a particle of mass m moving in a circular path at a distance r from the axis, 2 I mr  and v r   . For a uniform sphere of mass M and radius R and an axis through its center, 2 2 5 I MR  . The earth has mass 24 E m   5.97 10 kg , radius 6 E R   6.38 10 m and orbit radius 11 r   1.50 10 m . The earth completes one rotation on its axis in 24 h 86,400 s  and one orbit in 7 1 y 3.156 10 s   . Execute: (a) 2 24 11 2 40 2 7 2 rad (5.97 10 kg)(1.50 10 m) 2.67 10 kg m /s 3.156 10 s L I mr z z z                   . The radius of the earth is much less than its orbit radius, so it is very reasonable to model it as a particle for this calculation. (b)   2 2 2 24 6 2 33 2 5 5 2 rad (5.97 10 kg)(6.38 10 m) 7.07 10 kg m /s 86,400 s L I MR z z                  Evaluate: The angular momentum associated with each of these motions is very large. Note that the orbital angular momentum is over 7 orders of magnitude greater than its rotational angular momentum. In fact, most of the total angular momentum of the solar system comes from the sum of the orbital angular momenta of its planets.

orangejellyfish781

Consider an L-R circuit as shown in the figure.Thebattery provides 12.0 V of voltage. The inductor hasinductanceL, and the resistor has resistanceR= 150 Î©. The switch is initiallyopen as shown. Attime t = 0, the switch is closed. At time t aftert = 0 the currentI(t) flows through the circuit as indicated in thefigure.

What is the voltagechange V_r(t)across the resistor at time tafter t = 0?

(correct answer b)

What is the voltage changeV_ind(t)across the inductor at time t after t = 0? In thefollowingexpressions, ÎI/Ît denotes the time rate of change ofthecurrent in the inductor at time t?

(correct answer d)

After the switch is closed, thecurrentin the circuit grows over time approaching a constant value.Ingeneral, at time t after a voltage source is connected to anL-Rcircuit, the current I(t) in the circuit is given bytheexpression

I(t)=V/R(1-e^-t/Ï)

where V is the voltage provided bythebattery, R is the resistance of the resistor, and Ï is thetimeconstant characteristic of the circuit.

What is the current I(Ï) flowing inthecircuit shown in the figure at one time constant after t =0?

Express your answer in amperes.

Hint: To find the current inthecircuit after one time constant, use the expression forI(t)given in the problem, with t equal to Ï. Do not try todeterminethe time constant for this circuit. You would not be ableto do sobecause the inductance L is unknown. Instead, afterplugging in thenumerical values for V and R appropriatefor this circuit andsubstituting Ï for time, simplifyyour expression to get a numericalvalue for the current.