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Sergio [31]
3 years ago
8

Water falls without splashing at a rate of 0.370 l/s from a height of 2.90 m into a 0.690-kg bucket on a scale. if the bucket is

originally empty, what does the scale read 3.90 s after water starts to accumulate in it?
Physics
1 answer:
raketka [301]3 years ago
8 0
<span>R = rate of flow = 0.370 L/s H = height = 2.9 m T= time = 3.9 s V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2 G value = 9.8 m/s2 Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N Wa = weight of accumulated water after 3.9 s Fi = force of impact of water on the bucket S = reading on the scale = Wa + Wb + Fi mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg Therefore, Wa = 1.443 x 9.8 = 14.1414 N Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt) = 0.37 x 7.539 = 2.78943 N S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>
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The lowest point in Death Valley is 85 m below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 m.What is th
mario62 [17]

Answer:

\Delta E=2.87\times 10^6\ J

Explanation:

It is given that,

Depth of Death valley is 85 m below sea level, h_i=-85\ m

The summit of nearby Mt. Whitney has an elevation of 4420 m, h_f=4420\ m

Mass of the hiker, m = 65 kg

We need to find the change in potential energy. It is given by :

\Delta E=mg(h_f-h_i)

\Delta E=65\times 9.8(4420-(-85))

\Delta E=2869685\ J

or

\Delta E=2.87\times 10^6\ J

So, the change in potential energy of the hiker is 2.87\times 10^6\ J. Hence, this is the required solution.

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3 years ago
Suppose your car is on a 5% grade, meaning that for every 100 m you travel along the road you raise or lower only 5 m in elevati
kvv77 [185]

Answer:

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8 0
3 years ago
Read 2 more answers
A rubber balloon is filled with 1 L of air at 1 atm and 300 K and is then put into a cryogenic refrigerator at 100 K. The rubber
Kipish [7]

Answer:

The correct answers are

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to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem

The given variables are as follows

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V2 = Unknown

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let us assume that the balloon is perfectly elastic

At 300K the balloon is filled and it stretches to maintain 1 atmosphere

at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,

For (i) since we know that the pressure of the balloon is constant

by Charles Law V1/T1 =V2/T2

or V2 = (V1/T1)×T2 =\frac{1L}{300K}× 100K= \frac{1}{3} × L = L/3 hence the correct answer to (i) is 1/3L

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3 years ago
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olga_2 [115]

Answer:

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