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Liula [17]
1 year ago
5

Burt and ned paddle their canoe 10 miles upstream and 5 miles downstream in 7 hours. if they can paddle 4 mph in calm water, how

fast is the current?
Physics
1 answer:
mel-nik [20]1 year ago
6 0

The current flow is 2.4mph.

To find the answer, we have to know about the streamline flow.

<h3>How to find the rate of flow of current?</h3>

The boat's speed will reduce by V miles per hour if it is moving upstream because the current, which moves at a speed of V miles per hour, will be pressing against it. The boat moves downstream at a pace of Vp-V miles per hour as a result. However, if the boat is moving downstream, the current will be pushing it faster, increasing the boat's speed by V miles per hour. The boat's downstream speed as a result is Vp+V miles per hour.

where; Vp is the paddling speed in calm water.

  • It is given that, upstream distance is 10 miles, and downstream is 5 miles in 7 hours.
  • Thus, the time taken for downstream is,

                     t_d=\frac{S_d}{v_d}=\frac{5}{4+V}  , where; S is distance and v is speed.

  • Similarly, the time taken for upstream is,

                           t_u=\frac{S_u}{v_u}=\frac{10}{4-V}

  • The total time taken is,

                t=7hours=t_d+t_u\\                  

  • From this, the velocity of current is,

                       \frac{5}{4+V}+\frac{10}{4-V} =7\\\\V=2.4mph

Thus, we can conclude that, the current flow is 2.4mph.

brainly.com/question/25829296

#SPJ4

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Answer:L=109.16 m

Explanation:

Given

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mass flow rate of cold fluid \dot{m_c}=1.2 kg/s

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Let final Temperature be T

mass flow rate of geothermal water \dot{m_h}=2 kg/s

diameter of inner wall d_i=1.5 cm

U_{overall}=640 W/m^2K

specific heat of water c=4.18 kJ/kg-K

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)

2\times (160-T)=1.2\times (80-20)

160-T=36

T=124^{\circ}C

As heat exchanger is counter flow therefore

\Delta T_1=160-80=80^{\circ}C

\Delta T_2=124-20=104^{\circ}C

LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}

LMTD=\frac{80-104}{\ln \frac{80}{104}}

LMTD=91.49^{\circ}C

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

\dot{m_c}c(80-20)=U\cdot A\cdot (LMTD)

A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2

A=\pi DL=5.144

L=\frac{5.144}{\pi \times 0.015}

L=109.16 m

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An object is 12 m long, 0.65 m wide, and 13 cm high. Calculate the volume of this object.
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Mathematically the volume of this body is given as

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Note: The value given for the height was in centimeters, so it was transformed to meters.

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