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Ket [755]
2 years ago
14

A river has a steady speed of vs. A student swims upstream a distance d and back to the starting point. (a) If the student can s

wim at a speed of v in still water, how much time tup does it take the student to swim upstream a distance d? Express the answer in terms of d, v, and vs. tup = Incorrect: Your answer is incorrect. (b) Using the same variables, how much time tdown does it takes to swim back downstream to the starting point? tdown = Incorrect: Your answer is incorrect. Your answer cannot be understood or graded. More Information
Physics
1 answer:
andre [41]2 years ago
6 0

he speed of the student relative to shore is

v_ up = v- vs

v _down = v+ vs

The time required to travel distance d upstream
is

t_up = d/ v_up = d/ v- vs

(2)

The time required to swim the same distance d downstream is

t_down = d/ v_down = d/ v+ vs

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Please post in English so i or someone else can help you.
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2 years ago
Ever tried to stop a 150-pound (68 kg) cannonball fired towards you at 30 mph (48 km/hr.)? No, probably not. But you may have tr
alex41 [277]
The two situations are similar because in both you are trying to minimize the damage and make the best out of a bad situation
8 0
2 years ago
A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro
MArishka [77]

Answer:

Resistance = 3.35*10^{-4} Ω

Explanation:

Since resistance R = ρ\frac{L}{A}

whereas \rho(x) = a + bx^2

resistivity is given for two ends. At the left end resistivity is 2.25* 10^{-8} whereas x at the left end will be 0 as distance is zero. Thus

2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a

At the right end x will be equal to the length of the rod, so x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25}  = b\\b = 2.77 *10^{-8}

Thus resistance will be R = ρ\frac{L}{A}

where A = π r^2

so,

R = \frac{8.50*10^{-8} * 1.50}{3.14*(1.10*10^{-2})^2} \\R=3.35 * 10 ^{-4}

6 0
2 years ago
An underground tunnel has two openings, with one opening a few meters higher than the other. If air moves past the higher openin
klasskru [66]

Answer:

There would be a pressure drop in the direction of the higher opening. This will force air to move in from the lower opening and force it to leave through the higher opening. This will create a convectional movement of air, cooling and ventilating the tunnel.

Explanation:

This is in accordance with bernoulli's law of fluid flow which states that the pressure exerted by a moving fluid is lesser than it would exert if it were at rest.

6 0
2 years ago
A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b
trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}-------(1)

substitute P_a_t_m for P_1, (P_a_t_m +pgh) for P_2

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}

\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4

substitute

P_a_t_m for P_3, P_a_t_m for P_4

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

\sqrt{64.4(20-h)} for V₂

\sqrt{64.4h} for V₄

A₂V₂ = A₃V₃

A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft

Therefore ,height of the water rise in tank is 10ft

3 0
2 years ago
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