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yaroslaw [1]
3 years ago
15

The water in a river flows uniformly at a constant speed of 2.27 m/s between parallel banks 69.3 m apart. You are to deliver a p

ackage directly across the river, but you can swim only at 1.58 m/s. (a) If you choose to minimize the time you spend in the water, in what direction should you head? ° from the direction of the stream (b) How far downstream will you be carried? m (c) If you choose to minimize the distance downstream that the river carries you, in what direction should you head? ° from the direction of the stream (d) How far downstream will you be carried?

Physics
1 answer:
Tomtit [17]3 years ago
3 0

Answer: Got it!

Explanation: The water in a river flows uniformly at a constant speed of 2.50m/s between two parallel banks 80.0m apart. You are to deliver a package directly across the river, but you can only swim at 1.5m/s.

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Suppose you have two magnets. Magnet A doesn't have its poles labeled, but Magnet B does have a clearly labeled north and south
Levart [38]
<span>The right answer here would be option C - the side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole. This is due to the magnetic rule of opposites attracting.</span>
4 0
3 years ago
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A particle of mass m = 4.5 kg has velocity of v = 7 i m/s, when it is at the origin (0,0). Determine the z-component of the angu
Dima020 [189]

Answer:

Explanation:

Momentum P = Mass x Velocity

M = 4.5kg

V = 7m/s

4.5kg x 7m/s

= 31.5xkgm/s

L=IW( Angular momentum) at stationary origin (0,0)

I = 1/2 x Mr^2

L = 1/2 x 4.5x 31.5

L = 70.8kgm/s

At stationary point, (0,0) No coordinate exist

4 0
3 years ago
What is 34 + (5) × (1.2465) written with the correct number of significant figures?
bonufazy [111]

Answer: 40

Explanation:

= 34 + 5 * 1.2465

= ‭40.2325‬

= 40

The number of significant figures in the answer should be the same as the number with the least number of significant figures that any of the digits in the equation have.

32 has 2 significant figures so the answer has to be 2 significant figures which is 40.

7 0
3 years ago
A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d
prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance  

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

V_{t}= V_{o}e^{\frac{-t}{R*C} }

Voltage in this case is the energy dissipated so

E_{t}= E_{o}e^{\frac{-t}{R*C} }

\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }

\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }

Using the equation to find capacitance

ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }

C= 5.32x10^{-6} F

C= 5.32 uF because u is the symbol for micro that is equal to 10^{-6}

8 0
3 years ago
A cabbie is trying to stop when he notices a fare is whistling them over. The
liberstina [14]
  • K.E=18750J
  • Mass=m=2100kg
  • Velocity=v

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2

\\ \sf\longmapsto 18750=1050v^2

\\ \sf\longmapsto v^2=\dfrac{18750}{1050}

\\ \sf\longmapsto v^2=17.85m^2

\\ \sf\longmapsto v=\sqrt{17.85}

\\ \sf\longmapsto v=4.1m/s

7 0
3 years ago
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