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Andrews [41]
3 years ago
15

A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 66 m at t2 = 6.0 s .

Answer the following by considering ratios, without computing the car's velocity.(Express your answer to two significant figures and include the appropriate units.) 1.What is the car's position at t = 3.0 s ? 2.What will be its position at t = 24 s ?
Physics
1 answer:
UNO [17]3 years ago
7 0

Answer: 1. 33, 2. 264

Explanation: 66m= 6s so, to find the position at 3s you just need to take 66/2 = 33m cause 3 is half of 6. & for 2 you will take 66x4= 264m cause it took 4s multiply by the original 6s to get 24s. Answer: 1 is 33m and 2 is 264m

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A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate
yaroslaw [1]

Answer:

The  maximum error is  \Delta  \eta  = 2032.9

Explanation:

From the question we are told that

     The length  is  l  =  1\ m

      The radius is  r =  0.002 \pm  0.0002 \ m

        The pressure is  P  =  4 *10^{5} \ \pm 1750

        The  rate  is  v =  0.5*10^{-9} \ m^3 /t

       The viscosity is  \eta  =  \frac{\pi}{8} * \frac{P *  r^4}{v}

The error in the viscosity is mathematically represented  as

       \Delta  \eta  = | \frac{\delta \eta}{\delta P}| *  \Delta  P   +    |\frac{\delta \eta}{\delta r} |*  \Delta  r +  |\frac{\delta \eta}{\delta v} |*  \Delta  v

   Where  \frac{\delta \eta }{\delta P} =  \frac{\pi}{8} *  \frac{r^4}{v}

and         \frac{\delta \eta }{\delta r} =  \frac{\pi}{8} *  \frac{4* Pr^3}{v}

and          \frac{\delta \eta }{\delta v} =  - \frac{\pi}{8} *  \frac{Pr^4}{v^2}

So  

             \Delta  \eta  = \frac{\pi}{8} [ |\frac{r^4}{v}  | *  \Delta  P   +    | \frac{4 *  P * r^3}{v}  |*  \Delta  r +  |-\frac{P* r^4}{v^2}  |*  \Delta  v]

substituting values

            \Delta  \eta  = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}}  | *  1750   +    | \frac{4 *  4 *10^{5} * (0.002)^3}{0.5*10^{-9}}  |*  0.0002 +  |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2}  |*  0 ]

  \Delta  \eta  = \frac{\pi}{8} [56  +  5120 ]

   \Delta  \eta  = 647 \pi

    \Delta  \eta  = 2032.9

4 0
3 years ago
Gravity on the surface of the moon is only 1/6 as strong as gravity on the Earth. What is the weight of a 19 kg object on the Ea
Dahasolnce [82]

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Answer:

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General Formulas and Concepts:

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Explanation:

<u>Step 1: Define</u>

6.53 mol Mn

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1 mol Mn = 54.94 g Mn

<u>Step 3: Dimensional Analysis</u>

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<u>Step 4: Simplify</u>

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