Wh Question

two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 28 m = (12 m/s)t + This yields aB = – 2.5 m/s2. (b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB ? The requirement is 20 + 2t = 12t + 1 2 1 2 aB t2 where t = 4.0 s . aB t2 where aB = –5/2. There are two…

two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 28 m = (12 m/s)t + This yields aB = – 2.5 m/s2. (b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB ? The requirement is 20 + 2t = 12t + 1 2 1 2 aB t2 where t = 4.0 s . aB t2 where aB = –5/2. There are two…

two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 28 m = (12 m/s)t + This yields aB = – 2.5 m/s2. (b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB ? The requirement is 20 + 2t = 12t + 1 2 1 2 aB t2 where t = 4.0 s . aB t2 where aB = –5/2. There are two…

two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 28 m = (12 m/s)t + This yields aB = – 2.5 m/s2. (b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB ? The requirement is 20 + 2t = 12t + 1 2 1 2 aB t2 where t = 4.0 s . aB t2 where aB = –5/2. There are two…

two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 28 m = (12 m/s)t + This yields aB = – 2.5 m/s2. (b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB ? The requirement is 20 + 2t = 12t + 1 2 1 2 aB t2 where t = 4.0 s . aB t2 where aB = –5/2. There are two…

two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 28 m = (12 m/s)t + This yields aB = – 2.5 m/s2. (b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB ? The requirement is 20 + 2t = 12t + 1 2 1 2 aB t2 where t = 4.0 s . aB t2 where aB = –5/2. There are two…

questions: How much? How fast? How expensive? The question of how much is somewhat open ended. If the capacity is there, applications will likely be developed to use it. The question of how fast is, in a sense, easier to answer. To achieve greatest performance, the memory must be able to keep up with the processor. That is, as the processor is executing instructions, we would not want it to have to pause waiting for instructions or operands. The final question must also be…

pergunta, é uma afirmação: “Você tem que partir.” SNMs também são usados na ASL para indicar as negativas (“Você não tem que partir”), sentenças relativas (“O garoto que partiu é meu filho”), condicionais (“Se você partir, eu vou chorar”), perguntas wh- (“Onde você tem que ir?”) e perguntas retóricas (“Eu tenho que partir agora. Por quê? Porque eu tenho outro encontro do outro lado da cidade.”) em que não se espera que o interlocutor responda à pergunta “por quê?”. Sinais…

two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 28 m = (12 m/s)t + This yields aB = – 2.5 m/s2. (b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB ? The requirement is 20 + 2t = 12t + 1 2 1 2 aB t2 where t = 4.0 s . aB t2 where aB = –5/2. There are two…

two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 28 m = (12 m/s)t + This yields aB = – 2.5 m/s2. (b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB ? The requirement is 20 + 2t = 12t + 1 2 1 2 aB t2 where t = 4.0 s . aB t2 where aB = –5/2. There are two…