A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo after 0.0183 s. How far has the reflecting object moved? (Speed of sound = 343 m/s) (Unit = m)​

Distance of the object is 8.6 mDistance of the object is 8.6 mExplanation:Distance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6tDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature isDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree CDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree Cso we haveDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree Cso we havev = 332 + 0.6(20)v=332+0.6(20)Distance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree Cso we havev = 332 + 0.6(20)v=332+0.6(20)v = 344 m/sv=344m/sDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree Cso we havev = 332 + 0.6(20)v=332+0.6(20)v = 344 m/sv=344m/snow we know that bat heard the echo of sound in 0.05 sDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree Cso we havev = 332 + 0.6(20)v=332+0.6(20)v = 344 m/sv=344m/snow we know that bat heard the echo of sound in 0.05 sso the to and fro distance of the object is d + d = 2dDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree Cso we havev = 332 + 0.6(20)v=332+0.6(20)v = 344 m/sv=344m/snow we know that bat heard the echo of sound in 0.05 sso the to and fro distance of the object is d + d = 2dso we haveDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree Cso we havev = 332 + 0.6(20)v=332+0.6(20)v = 344 m/sv=344m/snow we know that bat heard the echo of sound in 0.05 sso the to and fro distance of the object is d + d = 2dso we have2 d = v\times t2d=v×tDistance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree Cso we havev = 332 + 0.6(20)v=332+0.6(20)v = 344 m/sv=344m/snow we know that bat heard the echo of sound in 0.05 sso the to and fro distance of the object is d + d = 2dso we have2 d = v\times t2d=v×t2d = 344(0.05)2d=344(0.05)Distance of the object is 8.6 mExplanation:As we know that the speed of sound at t degree C is given asv = 332 + 0.6 tv=332+0.6there we know that the temperature ist = 20 degree Cso we havev = 332 + 0.6(20)v=332+0.6(20)v = 344 m/sv=344m/snow we know that bat heard the echo of sound in 0.05 sso the to and fro distance of the object is d + d = 2dso we have2 d = v\times t2d=v×t2d = 344(0.05)2d=344(0.05)d = 8.6 md=8.6m

The sample with the lowest specific heat capacity will have the greatest temperature increase. in this case, it would be platinum.