A 2.50-mH inductor has a changing current from 0. to 12.00 A in 4.00 msec. What

Question

A 2.50-mH inductor has a changing current from 0. to 12.00 A in 4.00 msec. What emf is induced? a. 2.00 mV b. 7.20 mV c. 55.0 mV d. 2.50 V e. 7.50 V Which wavelength of light would be considered soft (low energy) gamma rays? a. 10 mm b. 10 mu m c. 10 nm d. 10 pm e. 10 ym A radio antenna is typically one quarter of a wavelength in length. How long is the typical antenna used to receive or send WNYC's signal at 93.9 MHz? a. 6.0 nm b. 12. mu m c. 80 cm d. 12.0 m e. 120 m You have a concave mirror with focal length 20 cm. How far from the mirror should you place an object to get a virtual image 15 cm from the mirror? a. 1.2 cm b. 8.6 cm c. 15. cm d. 20. cm e. 45. cm For the situation in problem 13, how is the image described? a. Virtual, upright, 75% larger b. Virtual, upright, 75% smaller c. Real, upright, 50% larger d. Real, inverted, 50% larger e. Virtual, inverted, 50% larger f. Virtual, inverted, 50% smaller g. Virtual, upright, 25% smaller

Explanation / Answer

Q10.

emf induced=inductor*rate of change of current

=2.5*0.001*(12-0)/(4*0.001)=7.5 volts

hence option e is correct.

Q11.

gamma rays are the highest energy EM raditation and have energy values are greater than 100 keV.

corresponding wavelength fall below 10 pm.

hence option d is correct.

Q12.

frequency=93.9 MHz

speed=3*10^8 m/s

then wavelength=speed/frequency=3.194 m

so radio antenna=(1/4)*3.194=0.7987 m

closest option is option c .

hence option c is correct.

Q13.

let object distance be u.
image distance =v=15 cm
focal length=f=-20 cm

using mirror formula:

(1/v)+(1/u)=1/f

==>u=-8.57 cm

so the object should be placed at a distance of 8.57 cm in front of the mirror

hence option b is correct.

Q14.

magnificiation=image height/object height=-image distance/object distance

==>image height/object height=-15/(-8.57)=1.75

so the image is virtual, upright, and 75% larger

hence option a is correct.

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