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NeX [460]
3 years ago
14

Can you check this? Samantha swam upstream for some distance in one hour. She then swam downstream the same river for the same d

istance in only 12 minutes. If the river flows at 4 mph, how fast can Samantha swim in still water?

Physics
2 answers:
Solnce55 [7]3 years ago
8 0

Choose the most logical value for the variable to represent. Let x= Kelli's swimming speed in still water


lorasvet [3.4K]3 years ago
4 0
So her speed in still water will be 6mph :)

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If the president vetoes a bill, it can still become law If congress overrides the presidents veto with ___ majority
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Which of the following are ways to improve muscle endurance:
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ASAP
KatRina [158]

Answer:

<h3>The answer is 11 mL</h3>

Explanation:

To find the volume of the object we use the formula

volume of object = final volume of water - initial volume of water

From the question

final volume of water = 86 mL

initial volume of water = 75 mL

So we have

volume of object = 86 - 75

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<h3>11 mL</h3>

Hope this helps you

4 0
3 years ago
To heat the house, the boiler transfers 15 MJ of energy in 10 minutes.
Harrizon [31]

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8 0
2 years ago
Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point
RideAnS [48]

Answer:

a) 2.33 m/s

b) 5.21 m/s

c) 882 m³

Explanation:

Using the concept of continuity equation

for flow through pipes

A_{1}\times V_{1} = A_{2}\times V_{2}

Where,

A = Area of cross-section

V = Velocity of fluid at the particular cross-section

given:

A_{1} = 0.070 m^{2}

V_{1} = 3.50 m/s

a) A_{2} = 0.105 m^{2}

substituting the values in the continuity equation, we get

0.070\times 3.50 = 0.105\times V_{2}

or

V_{2} = \frac{0.070\times 3.5}{0.150}m/s

or

V_{2} = 2.33m/s

b) A_{2} = 0.047 m^{2}

substituting the values in the continuity equation, we get

0.070\times 3.50 = 0.047\times V_{2}

or

V_{2} = \frac{0.070\times 3.5}{0.047}m/s

or

V_{2} = 5.21m/s

c) we have,

DischargeQ = Area (A)\times Velocity(V)

thus from the given value, we get

Q = 0.070m^{2}\times 3.5m/s\

Q = 0.245 m^{3}/s

Also,

DischargeQ = \frac{volume}{time}

given time = 1 hour = 1 ×3600 seconds

substituting the value of discharge and time in the above equation, we get

0.245m^{3}/s = \frac{volume}{3600s}

or

0.245m^{3}/s\times 3600 = Volume

volume of flow = 882 m^{3}

8 0
3 years ago
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