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kumpel [21]
3 years ago
10

A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat

er. Find the true velocity of the swimmer as a vector. (Assume that the i vector points east, and the j vector points north.)

Physics
1 answer:
Dafna11 [192]3 years ago
3 0

Answer:

<u>velocity of swimmer relative to ground = 3 i -5 j</u>

Explanation:

  • To cross a river the swimmer swims relative to river in perpendicular direction.

Velocity of river = -5 j (south)

Velocity of swimmer relative to river = 3 i(north)

So

<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>

Velocity of swimmer relative to ground = 3 i -5 j

So magnitude of total velocity is \sqrt{3^2+(-5)^2} =\sqrt{9+25} = \sqrt{34}

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Janelle stands on a balcony, two stories above Michael. She throws one ball straight up and one ball straight down, but both wit
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Answer:

Both balls have the same speed.

Explanation:

Janelle throws the two balls from the same height, with the same speed. Both balls will have the same potential and kinetic energy. Energy must be conserved. When the balls pass Michael, again they must have the same potential and kinetic energy.

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Which explanation of the solar system best fits the observations of the planets and how they orbit the sun?
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The dust and gas accumulate to form a solar nebula, which later on creates the star and the planets.
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A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on M
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3 years ago
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

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