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kondor19780726 [428]
3 years ago
15

A boat moves through the water of a river at 4.72 m/s relative to the water, regardless of the boat’s direction. If the current

is flowing at 3.86 m/s, how long does it take the boat to complete a trip consisting of a 297 m displacement downstream followed by a 389 m displacement upstream? Answer in units of s.
Physics
2 answers:
Tems11 [23]3 years ago
7 0

Answer:

486.95s

Explanation:

Let the velocity of the boat be v_b and that of the river be v_r.

When the boat is moving downstream, the resultant velocity is given by;

v=v_b+v_r\\Given;\\v_b=4.72m/s\\v_r=3.86m/s\\therefore\\v=4.72m/s+3.86m/s=8.58m/s

Recall that

velocity=\frac{dispacement}{time}.............(1)

Given; downstream displacement, d = 297m. Therefore by equation (1);

8.58=\frac{297}{t_1}\\t_1=\frac{297}{8.58}\\t_1=34.62s

where t_1 is the time taken to travel downstream displacement.

When the boat is moving upstream it is moving directly against the river current, hence its resultant velocity in this case becomes;

v=v_b-v_r\\v=4.72-3.86=0.86m/s

The time taken upstream is then calculated as follows given that the upstream displacement is 389m, according to equation (1);

0.86=\frac{389}{t_2}\\

therefore t_2=\frac{389}{0.86}=452.326s

Hence the total time taken by the boat to complete its trip as specified is given as follows;

t_{total}=t_1+t_2= 34.62+452.326\\t_{total}=486.95s

Setler [38]3 years ago
5 0

Answer:

Total time taken by boat to cover all the displacement = 452.3254 sec

Explanation:

We have given speed of boat = 4.72 m/sec

Speed of current flowing = 3.86 m./sec

In downstream relative speed of boat = speed of boat + speed of current flowing = 4.72+3.86 = 8.58 m/sec

Distance traveled downstream = 297 m

So time taken by boat to travel 297 m downstream t_1=\frac{297}{8.58}=34.615sec

In upstream relative speed of speed = speed of boat - speed of current flowing = 4.72 -3.86 = 0.86 m/sec

Distance traveled in upstream = 389 m

So time taken to travel 289 m upstream t_2=\frac{389}{0.86}=452.325sec

So total time t=t_1+t_2=34.615+452.325486.94 sec

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<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

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                       M = Mass of body 1

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                       r = Distance between them

Here we have

                 M = Mass of Sun = 1.99×10³⁰ kg

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Substituting

                   F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m

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Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

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