Answered: Figure P30.4 n turns/m R…

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Physics

Figure P30.4 n turns/m R coc0000ço000000000000 N turns

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter9: Energy In Nonisolated Systems

Section: Chapter Questions

Problem 49PQ

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A a -> Two water Separated from A ond B motors are Points at the Some time. Tw o what is the shortest distonce betwe en the bot ?
CASE STUDY In Example 34.6 (page 1111), we imagined equipping 1950DA, an asteroid on a collision course with the Earth, with a solar sail in hopes of ejecting it from the solar system. We found that the enormous size required for the solar sail makes the plan impossible at this time. Of course, there is no need to eject such an object from the solar system: we only need to change the orbit. A much more pressing problem is Apophis, a 300-m asteroid that may be on a collision course with the Earth and is due to come by on April 13, 2029. It is unlikely to hit the Earth on that pass, but it will return again in 2036. If Apophis passes through a 600-m keyhole on its 2029 pass, it is expected to hit the Earth in 2036. causing great damage. There are plans to deflect Apophis when it comes by in 2029. For example, we could hit it with a 10- to 150-kg impactor accelerated by a solar sail. The impactor is launched from the Earth to start orbiting the Sun in the same direction as the Earth and Apophis. The idea is to use a solar sail to accelerate the impactor so that it reverses direction and collides head-on with Apophis at 8090 km/s and thereby keeps Apophis out of the keyhole. Consider the momentum in the impactors orbit (Fig. P34.75) when the solar sail makes an angle of = 60 with the tangent to its orbit. Current solar sails may be about 40 m on a side, but the hope is to construct some that are about 160 m on a side. Estimate the impactors tangential acceleration when it is about 1 AU from the Sun. Keep in mind that the sail is neither a perfect absorber nor a perfect reflector, and a heavier impactor would presumably be equipped with a larger sail. Dont be surprised by what may seem like a very small acceleration. FIGURE P34.75
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1. A 11.8 ?F capacitor is charged by a 25.0 V battery through a resistance R. The capacitor reaches a potential difference of 4.00 V at a time 3.00 s after charging begins. Find R. ________ kΩ 2.A proton moves perpendicular to a uniform magnetic field at a speed of 1.20 107 m/s and experiences an acceleration of 1.40 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field. magnitude _______ PLEASE ANSWER BOTH
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The figure below shows a conducting rod sliding along a pair of conducting rails. The conducting rails have an angle of inclination of θ=30 degrees. There is a resistor at the top of the ramp that connects the two conducting rails R=2.3Ω. The mass of the rod is 0.42 kg. The rod starts from rest at the top of the ramp at time t=0. The rails have negligible resistance and friction, and are separated by a distance L=15.7 m. There is a constant, vertically directed magnetic field of magnitude B=1.5 T. A) Find the emf induced in the rod as a function of its velocity down the rails. What is the emf when the velocity is 5.696E−03 m/s? B) What is the rod's terminal speed? C) When the rod moves at its terminal speed, what is the power dissipated in the resistor?
The figure below shows a conducting rod sliding along a pair of conducting rails. The conducting rails have an angle of inclination of θ=30 degrees. There is a resistor at the top of the ramp that connects the two conducting rails R=2.3Ω. The mass of the rod is 0.42 kg. The rod starts from rest at the top of the ramp at time t=0. The rails have negligible resistance and friction, and are separated by a distance L=15.7 m. There is a constant, vertically directed magnetic field of magnitude B=1.5 T. Find the emf induced in the rod as a function of its velocity down the rails. What is the emf when the velocity is 5.696E−03 m/s? .116 W What is the rod's terminal speed? 0.01138 m/s WHAT I NEED HELP WITH: A) When the rod moves at its terminal speed, what is the power dissipated in the resistor? For this I was using the equation P = V2/R and I got .00586 W which is wrong. What am I doing wrong?