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MrRissso [65]
3 years ago
15

Problem set 3 » (8) river swimming a swimmer heads directly across a river, swimming at her maximum speed of 1.30 m/s relative t

o the water. She arrives at a point 48.0 m downstream from the point directly across the river, 64.0 m wide. What is the speed of the river current?
Physics
1 answer:
jasenka [17]3 years ago
5 0

Velocity of swimmer across river = 1.30 m/s

Distance arrived downstream = 48 m

Width of river = 64 m

Time taken to cross river = \frac{Width of river}{Velocity across river}

                                          = \frac{64}{1.30} =49.23 s

Speed of river current = \frac{Distance arrived downstream}{Time taken to cross river}

                                     = \frac{48}{49.23} = 0.975 m/s

So, the river is flowing at a speed 0.975 m/s.

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Which happens to the magnetic field of a wire when you change the direction of the current in the wire? It becomes stronger. It
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Your neighbor is riding her bike around the block. When she slows down and turns a corner, what changes about her?.
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Answer:

nothing

Explanation:If you ride a bike around the block and return to the exact point where you started, your displacement is zero.

By definition, displacement involves changing an object from its original position. No matter how far or for how long a body moves, if it returns to the position it started from, it has not been displaced at all. This means that the body has zero displacements.

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Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the
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Answer:

T=1.54s

Explanation:

From the question we are told that:

Speed of Motor 1 \omega_1=825rpm=>2 \pi 13.75

Speed of Motor 2 \omega_1=786rpm=>2 \pi 13.1

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Frequency of Motor 1 f_1=13.75

Frequency of Motor 2  f_2= 13.1

Generally the equation for Time Elapsed is mathematically given by

T=\frac{1}{df}

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df=f_1-f_2

df=13.75-13.1

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T=\frac{1}{65}

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5 0
3 years ago
n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble
Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm  is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is    U_f -U_i = 16 *10^{-7} J

Explanation:

From the question we are told that

     The radius of the soap bubble  is  R_o =  7.45 \ cm =  \frac{7.45}{100} =  0.0745 \ m

      The potential of the soap bubble is  V_1  =307.0 V

      The new radius of the soap bubble  is R_1 =  0.19 * 7.45=1.4155\ cm = 0.014155 \ m

The initial electric potential is mathematically represented as

     U_i  = \frac{V_1^2 R_o }{2k }

The final  electric potential is mathematically represented as

    U_f  = \frac{V_2^2 R_1 }{2k }

The initial potential is mathematically represented as

     V_1 =  \frac{kQ}{R_o}

The final  potential is mathematically represented as

        V_2 =  \frac{kQ}{R_1}

Now  

         \frac{V_2}{V_1}  =  \frac{R_o}{R_1}

substituting values

        \frac{V_2}{V_1}  =  \frac{7.45}{1.4155} =   \frac{1}{0.19}

=>      V_2 =  \frac{V_1}{0.19}

    So

         U_f  = \frac{V_1^2 R_2 }{2k * 0.19^2}

Therefore

        U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }

       U_f -U_i =     \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]

where k is the coulomb's constant with value 9*10^{9} \  kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

       U_f -U_i =     \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]

       U_f -U_i = 16 *10^{-7} J

           

     

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