Use this formula to find your answer...
Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms
frequency (f)=1/( Time period).
Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.
Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.
Mechanical energy is the sum of kinetic energy and potential energy
Get a pull chord light switch installed in her bathroom by a qualified electrician asap. Meanwhile, keep hands as dry as possible, and try not to go near that switch until it's either been properly earthed, or whatever the problem actually is, and get qualified advice on what the problem is. Don't have wet feet either, and don't stand in puddles of water whilst operating - i she has to - the switch. 250V AC mains can be lethal, and at least painful.
we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m
<span>No sé una palabra que acaba de decir, ¿se puede decir en inglés por favor ???</span>
