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Mekhanik [1.2K]
2 years ago
9

The following times are given using metric prefixes on the base SI unit of time: the second. Rewrite them in scientific notation

without the prefix. For example, 47 Ts would be rewritten as 4.7×1013s. (a) 980 Ps; (b) 980 fs; (c) 17 ns; (d) 577μs.

Physics
2 answers:
Leno4ka [110]2 years ago
8 0

Answer: a=9.8*10^-10s

b=9.8*10^-13s

c=1.7*10^-8s

d=5.57*10^-4s

Explanation:

a) given 980 ps

Expected answer is 980 * 10^-12

Therefore, 980ps = 9.8*10^-10s

b) given 980 fs

Expected answer is 980 * 10^-15

Therefore, 980fs = 9.8*10^-13s

c) given 17 ns

Expected answer is 17 * 10^-9

Therefore, 17ns = 1.7*10^-8s

d) given 577 μs

Expected answer is 577 * 10^-6

Therefore, 577μs = 5.57*10^-4s

a=9.8*10^-10s

b=9.8*10^-13s

c=1.7*10^-8s

d=5.57*10^-4s

Evgesh-ka [11]2 years ago
5 0

Explanation:

Below is an attachment containing the solution.

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Explanation:

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x=10 m  The horizontal distance of the water ballon

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We need to find the initial velocity V_{o} at which the water ballon was thrown, and we can find it by the following equation:

x=V_{o}cos \theta T (1)

Where T=2t is the total time the water ballon is on air

On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when V=0 and the time t is half the time T it takes the complete parabolic path.

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Isolating t:

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V_{o}=\sqrt{\frac{(10 m)(9.8 m/s^{2})}{-2 cos(35\°)}} (7)

Finally:

V_{o}=7.734 m/s

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