Rh Quick Case

55780644887275 × 10−3 s which should now be rounded to 3.88 × 108 rotations since the time-interval was specified in the problem to three significant figures. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or 1 × 106 = 1 rotation t 1.55780644887275 × 10−3 s which yields the result t = 1557.80644887275 s (though students who do…

55780644887275 × 10−3 s which should now be rounded to 3.88 × 108 rotations since the time-interval was specified in the problem to three significant figures. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or 1 × 106 = 1 rotation t 1.55780644887275 × 10−3 s which yields the result t = 1557.80644887275 s (though students who do…

× 10−3 s which should now be rounded to 3.88 × 108 rotations since the time-interval was specified in the problem to three significant figures. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or 1 × 106 = 1 rotation t 1.55780644887275 × 10−3 s which yields the result t = 1557.80644887275 s (though students who do this…

communicate to each other using sub-audible sounds, as do male alligators seeking a mate. Similarly, operating in the super-audible register, there is the dog whistle, which we cannot hear, as well as the sonic pulses emitted by bats. In both of these cases, the frequencies are so high that we do not register them. The range of frequencies generally used in music covers a little more than seven octaves of sound, which is the general range covered by a concert grand piano. The ranges of…

and an average speed of 14.0 m s . b) The first stage of 240 the journey takes 8.0 mms = 30 s and the second stage of the journey takes (240 m 20 m s) = 12 s , so the time for the 480-m trip is 42 s, for an average speed of 11.4 m s . c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same. 2.8: From the expression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) 20.8 m − 5.60 m 5.60 m −0 20.8 m − 0 = 7.6…

and an average speed of 14.0 m s . b) The first stage of 240 the journey takes 8.0 mms = 30 s and the second stage of the journey takes (240 m 20 m s) = 12 s , so the time for the 480-m trip is 42 s, for an average speed of 11.4 m s . c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same. 2.8: From the expression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) 20.8 m − 5.60 m 5.60 m −0 20.8 m − 0 = 7.6…

55780644887275 × 10−3 s which should now be rounded to 3.88 × 108 rotations since the time-interval was specified in the problem to three significant figures. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or 1 × 106 = 1 rotation t 1.55780644887275 × 10−3 s which yields the result t = 1557.80644887275 s (though students who do…

55780644887275 × 10−3 s which should now be rounded to 3.88 × 108 rotations since the time-interval was specified in the problem to three significant figures. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or 1 × 106 = 1 rotation t 1.55780644887275 × 10−3 s which yields the result t = 1557.80644887275 s (though students who do…

and an average speed of 14.0 m s . b) The first stage of 240 the journey takes 8.0 mms = 30 s and the second stage of the journey takes (240 m 20 m s) = 12 s , so the time for the 480-m trip is 42 s, for an average speed of 11.4 m s . c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same. 2.8: From the expression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) 20.8 m − 5.60 m 5.60 m −0 20.8 m − 0 = 7.6…

an average speed of 14.0 m s . b) The first stage of 240 the journey takes 8.0 mms = 30 s and the second stage of the journey takes (240 m 20 m s) = 12 s , so the time for the 480-m trip is 42 s, for an average speed of 11.4 m s . c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same. 2.8: From the expression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) 20.8 m − 5.60 m 5.60 m −0 20.8 m − 0 = 7.6 m s