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Dmitriy789 [7]
3 years ago
6

A​ boat's crew rowed 33kilometers​ downstream, with the​ current, in 3hours. The return trip​ upstream, against the​ current, co

vered the same​ distance, but took 11hours. Find the​ crew's rowing rate in still water and the rate of the current.
Mathematics
1 answer:
algol [13]3 years ago
6 0

Answer:

The crew rowing rate in still water = 7km/hr

The rate of the current = 4km/hr

Step-by-step explanation:

The distance rowed downstream with the current = 33km

Time taken to row downstream = 3hrs

The distance rowed upstream against the current = 33km

Time taken to row upstream = 11hrs

Let the crew's rowing rate in still water = Xkm/hr

Let the rate of current = Ykm/hr

For the downstream movement, we will have (X+Y). This is because the boat rowed with the current

For the upstream movement, we will have (X-Y). This is because the boat rowed against the current

Downstream Speed = distance /time

= 33/3

= 11km

Upstream speed = distance /time

= 33/11

= 3km

Hence X+Y = 11 and X-Y= 3

Add both equations. We have

X+Y+X-Y = 11 + 3

2X = 14

X= 14/2

X= 7

put X= 7 in any of the equations

7-Y= 3

Y= 7-3

Y= 4

Hence the crew rowing rate in still water = 7km/hr and the rate of the current = 4km/hr

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Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




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