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Svetach [21]
3 years ago
7

Water flows through a cylindrical pipe of varying cross-section. The velocity is 3.0 m/s at a point where the pipe diameter is 1

.0 cm. What is the flow rate R
Physics
1 answer:
Setler79 [48]3 years ago
8 0

Answer:

The flow rate is  R =2.357 *10^{-4} \ m^3/s

Explanation:

From the question we are told that

    The velocity is v  =  3.0 \ m/s

   The  diameter of the pipe is  d =  1.0 \ cm  = 0.01 \ m

 

The  radius of the pipe is mathematically represented as

            r =  \frac{d}{2}

substituting values

            r =  \frac{0.01}{2}

           r =  0.005 \ m

The flow rate is mathematically represented as

       R  =  v  * A

Where is the cross-sectional area of the pipe which is mathematically evaluated as

      A   = \pi r^2

substituting values

      A   =  3.142 *  (0.005)^2

     A   = 7.855 *  10^{-5} \ m^2

So

    R  =  3.0  *  7.855 *10^{-5}

    R  =  2.357*10^{-4} \ m^3 /s

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Explanation:

Given that,

Current (I) = 2 mA

(Since 1 mA = 1 x 10^-3A

2 mA = 2 x 10^-3A)

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Time taken for a fully charged phone to die (T) = ?

Recall that the charge is the product of current and time taken.

i.e Q = I x T

130C = 2 x 10^-3A x T

T = 130C / (2 x 10^-3A)

T = 65000 seconds (time will be in seconds because seconds is the unit of time)

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6 0
3 years ago
A 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot
Molodets [167]

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

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Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

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