Un iman se mueve dentro de una bobina_ Considera os siguientes LGSOCaumentamos campo magnetico del imanel numero de vueltasvelocidaduainan ze mueveEn que casos {ugi...

Speedometer measures the speed of the car in (a) $\mathrm{m} \mathrm{s}^{-1}$ (b) $\mathrm{km} \mathrm{h}^{-1}$ (c) $\mathrm{cm} \mathrm{s}^{-1}$ (d) $\mathrm{km} \mathrm{min}^{-1}$

We're taking a look at this roading. Get equation here. What we have is our operator way function and some energy return. Once applied by the way function, we're also considering the how Morton operator for the harmonic oscillator, which is H bar H hot. Sorry. Equals negative H r squared over two mu. Second derivative two pi b squared mu X hot squared. So what? We have ISS the way function X e to the minus a x squad. What energy is be at one of two, um HP and so energy one of the two HB. Once he substitute in zero here. Now, we substitute in the way function on energy into our very fast equation. Then we simplify that damn where we determine that the ground state wave function is an Eigen function. And so therefore, two pi squad B squared mu minus h bar squared over two mu for a squared. Really cool zero. Then we solve for a which is hi. Be juice mass over H bar

According to the Doppler effect. Yeah, and frequency of light received is less than frequency of light produced by the sorts that is source is moving away. That is car is moving away from I don't know, no in second part. Here it is given frequency have to is cut to four point Jiro four kilohertz mhm and F is cut to 24.15 New cards. Yeah, according to Dr Effect Frequency receive actresses, but to have but minus you were policy. You can be lieutenants minus c have 10 minus f upon f Mm, the total door perceived yeah of frequency. After it's not changing frequency between jestrada between or not, um, frequency received by moving car Mhm so deep can like one plus x square. So one plus two access card to zero access card to mind yourself has given the problem. So have best minus every school too. Action to F two. So it is to be minus 2.2 Lourdes. Yeah. Mhm. No, you. It's going to my necessity. Our basement has ever part of a substitute The value three into 10 to the power meter per city to 20 upon Yeah, 24.15 10 to the power nine house. So to speak. After card, you will get 35.1 m per second. Okay? Yeah, there's one. Thanks for watching it.

Hi, everybody is somebody from the amplitude of vibration. And so we give And this for two conditions. And here we have the frequency given to us as to pie in the better by 60 and it's gonna be equal to two Pi 900 divided by 60 equals 94.245 and then we have our PM is gonna be AM radius times Your frequency squared equals 0.64 times 0.19 2942 times 6.25 divided by 12 because we wanted in feet. Okay, times 94.245 squared. This is gonna equal 5.7 for nine pounds. And for a we have X m equals P m divided by K minus the frequency squared. Um, And now we got Pete, um, equals K minus weight right away. Gravity times different C squared. And we could enter the values and to hear. So your ex, um, equals five 0.749 Divided by 57 60 minus 36. Divided by 32.2 times 94.24 squared. This is not equal to negative 0.0 Wine 378 and weaken times this by 12 inches to get inches of out of my feet. So then we get negative. 0.0 wine. 654 inches. Okay. And for the second part, it lets me right. Here we go. Be we have your CC is two k m. As your critical damp Ian is too Times square root of 500 me five thousands of 160 times. And the K M is your, um, spring constant 100 mass and need amassed 36 divided by 32.2. And let me just write down right at 32.2, we write the whole thing's. Apparently I cannot rate computer writing is a lot harder than handwriting tell you that. Okay. And this is very equal calculator of £160.496.6 seconds ahead of Eid feet. Okay. And your c is the equal. 0.55 c Time CC. So? So I'm gonna be equal to 0.55 times 1 60.496 Okay, on this is gonna eagle Teoh 8.8 to 73 pounds time Second divided by feet. Okay. And now we have X m equals he, um and we have K minus. Um, Did you reprint? See squared squared, plus C w we print see squared and you can enter the information that we have for it as 5.7 for nine, divided by 57 60 minus 36. Divided by 32.2 for the mass times 94 points, 94.24 squared. And this whole thing is squared. Remember, this whole thing is square rooted as well. Okay. And plus, we got 8.82 and actually one, wouldn't you? I told you that. 8.8 to 73 times 94.24 This thing is square rooted. And so this is going equal to 0.1352 24 beats and again, like above need two times this by 12 inches, divided by feet. And then we get 0.1622 inches. And that is your answer for part B. Okay, guys,

Hi. In the given problem here, this is not This is solved. And here this is east and this is rest. The West direction is coming out for pentacle to the plane of paper and east direction is into the plane. Off paper, the vertical antenna is shown like this. Suppose this arrow is representing vertical antenna. The direction off Earth's magnetic field is making an angle off 72 degree below the horizontal means. Suppose this is arts magnetic field. This whose components will be This will be the horizontal component and this will be the vertical it down work component. This is edge the horizontal component off what's Magritte will be. And this is the vertical component. This is angle of declination Delta, which has been given us 72 degree. Yeah, this is 72 degrees length off the antenna. This empty now has been given us 85 centimeter. Or we can say this is zero point 85 m the velocity with which this car is moving due east. This is the direction off motion off the car with the velocity V, which is given as 25 m per second do east sort it is obvious that this word Valentina will intercept on Lee the horizontal component off Earth's magnetic field. In the first part of the problem, we have to find which end off this antenna top end or the bottom and off the antenna will be at a higher potential. So so find it. We use Fleming's right and rule. So using blame ing's right and rule as the velocities towards east the horizontal component from south to north that is, towards left. So using right hand rule, it is obvious that the current will be flowing in upward direction. This is the direction of current using Fleming site and rule. We conclude that the current the induced current we'll pass through antenna in upward direction and, as we know, induced current do do emotional MF. All this rents from look do high potential, so it is obvious that the top end off the antenna top off the antenna will be at high potential holly for attention. And this is the answer for the first part off the problem. Now, in the second part of the problem, we have to find the magnitude off that emotional MF induced for which we will use simply the expression for emotionally IMF, which is given as the product of magnetic field with the velocity with the length and as here, this magnetic police only the horizontal home phone, and so we use edge for it. Then we then l for horizontal component. The expression will be be cause Delta and to be in tow help. So now, plugging in all known values for Earth's magnetic field, this is 5.9 into 10. Dish par minus five or sign off 72 degree into re means 25 m per second into 0.85 the length of the angina. So finally, this e m f here comes out to be 013 87 into 10 dish bar minus three volts. Or we can say this is 0.39 Milly Ward and this is the answer. The second answer for the given problem. Thank you