ch33 tủ tài liệu training

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ch33 tủ tài liệu training

I.In air. light travels at roughly c - 3.0 X 10s ni/s. Therefore, for t 1.0 ns. we have a distance ofd = Ct = (3.0 X10K m / s) (1.0 X10s) = 0.30 in.2(

ch33 tủ tài liệu training(a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm.(b)and the larger wavelength to be approximately 610 nm.(c)From Fig.

33-2 the wavelength at which the eye is most sensitive is about 555 nm.(d)Using the result in (c), we haveT = 3”xjo?/s = 5.41x10» Hz.Ã 555 nm(c) The p ch33 tủ tài liệu training

eriod is (5.41 X IO14 Hz) * = 1.85 X 10 15 s.3(a) The frequency of the radiation isc3.0x10" in/sJ /. (I.0x105)(6.4x106m)-4.7x10 ’Hz.(b) The period of

ch33 tủ tài liệu training

the radiation isT=-^ = -./■ 17x10 Hz= 212 S = 3min32 s.4Since _v. À, we find 4/ is equal toJsL = (lOxm^XO^OOxm’n,) = 7.49x 10’ Hz.X2 (632.8x10 m)25If/

I.In air. light travels at roughly c - 3.0 X 10s ni/s. Therefore, for t 1.0 ns. we have a distance ofd = Ct = (3.0 X10K m / s) (1.0 X10s) = 0.30 in.2(

ch33 tủ tài liệu trainingrent in the LC circuit of the generator. That is, f - \!2k\ LC , where c is the capacitance and L is the inductance. ThusX___=2k/lC c'The solution for

I. isX2(550X10 9 m)?__L =', . ?. = —7----- 7 . ----------:----T - 5.00 X1 O’2' H.4k Cc 4it2(17x 10"12 F)(2.998 X10* m/s)This is exceedingly small.6Th ch33 tủ tài liệu training

e emitted wavelength isX=£=2ne7Zc = 2k(2.998x 1 0' m/s) ự(o.253x 1 O'* H)(25.0x 10“12 F) = 4.74m.

I.In air. light travels at roughly c - 3.0 X 10s ni/s. Therefore, for t 1.0 ns. we have a distance ofd = Ct = (3.0 X10K m / s) (1.0 X10s) = 0.30 in.2(