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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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Essay about why people should not join a gang 300 word​
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Answer:

I don't know if it's fair for me to write an entire essay for you

but if you would like I can list some reasons you can incorporate into an essay.

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-If you are joining a gang you are most likely going to have a greater chance of being targeted, even if it seems like you would be protected.

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6 0
2 years ago
If the surface air pressure is 1000 mb and the pressure at the top of the atmosphere (75 km) is 0 mb, at what altitude would I f
Lana71 [14]

Answer: 5.5km

Explanation:

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Pressure decreases with increasing altitude. This is because at At higher altitudes, there are fewer air molecules above a the known or given surface than a similar surface at lower levels.

Pressure decreasing with higher altitudes also means that  air pressure decreases rapidly at lowerevels but more slowly at higher levels.

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8 0
3 years ago
A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i
mars1129 [50]

Complete Question

The complete question is  shown on the first uploaded image  

 

Answer:

The electric field at that point is  E = 7500 \ N/C

Explanation:

From the question we are told that  

       The  radius of the inner circle is r_i  =  0.80  \ m

        The  radius of the outer circle is  r_o  =  1.20 \ m

       The  charge on the spherical shell q_n  =  -500nC  = -500*10^{-9} \ C

      The magnitude of the point charge at the center is  q_c =  + 300 nC  =  + 300 * 10^{-9} \ C

        The  position we are considering is  x =  0.60 m  from the center

Generally  the  electric field  at the distance x =  0.60 m  from the center  is mathematically represented as

                 E =  \frac{k *  q_c   }{x^2}

substituting values  

                  E =  \frac{k *  q_c   }{x^2}

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

     substituting values

                  E =  \frac{9*10^9  *  300 *10^{-9}}{0.6^2}

                 E = 7500 \ N/C

7 0
3 years ago
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