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BlackZzzverrR [31]
2 years ago
12

A 432 g sample of 60/27Co has a decay constant of 4.14 x 10-9 s-1. How long will it take before only 1/3 of the original sample

remains?
Physics
1 answer:
Musya8 [376]2 years ago
7 0

Answer:

remain 1s60

Explanation:

I took away sample

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A startled armadillo leaps upward, rising 0.537 m in the first 0.220 s. (a) what is its initial speed as it leaves the ground? (
Elena L [17]
Let u = initial vertical velocity.

Assume that
g = 9.81 m/s²,
Wind resistance is ignored.

When t = 0.220 s, the height is h = 0.537 m. Therefore
0.537 m = (u m/s)*(0.220 s) - (1/2)*(9.81 m/s²)*(0.220 s)²
0.537 = 0.22u - 0.2372
u = 3.519 m/s

The upward velocity after 0.220 s is
v = 3.519 - 9.81*0.22 = 1.363 m/s

At maximum height, the upward velocity is zero. The maximum height, H, is given by
(3.519 m/s)² - 2*(9.81 m/s²)*(H m) = 0
12.3834 - 19.6H = 0
H = 0.632 m
It goes higher by 0.632 - 0.537 = 0.095 m

Answers:
(a) The initial speed is 3.519 m/s.
(b) The speed at  0.537 m height is 1.363 m/s.
(c) It goes higher by 0.095 m.

3 0
3 years ago
What is the potential difference when the current in a circuit is 5mA and resistance is 30 Ohms
Mashcka [7]

<h2>\bf{ \underline{Given:- }}</h2>

\sf• \: The \:  current \:  in \:  a \:  circuit \:  is  \: 5 \: amps.  \: and  \: resistance \:  is \:  30 \:  Ohms.

\\

<h2>\bf{ \underline{To \:  Find :- }}</h2>

\sf{• \:  The  \: Potential  \: Difference. }

\\

\huge\bf{ \underline{ Solution:- }}

\sf According  \: to  \: the  \: question,

\sf•  \: Current \:  (I) = 5  \: Amps.

\sf• \:  Resistance  \: (R) = 30 \:  Ω

\sf{Potential \:  difference  \: means  \: Voltage \: ( V).}

\sf{We \:  know \: that, }

\bf \red{ \bigstar{ \: V = IR }}

\rightarrow \sf V =5 \times 30

\rightarrow \sf V =150

\\

\sf \purple{Therefore, \:  the \:  potential  \: difference  \: is  \: 150  \: v \: .}

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