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3 years ago

13

Write a function digits() that accepts a non-negative integer argument n and returns the number of digits in it’s decimal repres

entation. For credit, you should implement this using a loop and repeated integer division; you should not use math.log(), math.log10(), or conversion to string

1 answer:

Katyanochek1 [597]3 years ago

3 0

Answer:

Explanation:

Let do this in python. We will utilize the while loop and keep on dividing by 10 until that number is less than 10. Along the loop we will count the number of looping process. Once the loop ends, we can output that as answer.

def digits(n):

n_digit = 1

while n >= 10:

n_digit += 1

n /= 10

return n_digit

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solniwko [45]

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3 years ago

Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What

Tresset [83]

Answer:

the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

Explanation:

Given that:

Initial Temperature $T_1$ = 15°C

Initial Pressure $P_1$ = 5 atm

Final Pressure $P_2$ = 1 atm

Data obtain from steam tables of saturated water at 15°C are as follows:

Specific volume v = 1.001 cm³/gm

The change in temperature = 2°C

Specific heat of water = 4.19 J/gm.K

volume expansivity β = 1.5 × 10⁻⁴ K⁻¹

The expression to determine the change in temperature can be given as :

$\delta \ T = \frac{-V (1- \beta \ T}{C_p} * \delta \ P ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$ $\delta \ T = \frac{-1.001 \frac{cm^3}{gm} (1- 1.5*10^{-4} \ K^{-1} )*2}{4.19 \ \frac{J}{gm.K}} *(5-1)atm ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$

Δ T = 0.093 K

Now; we can calculate the lost work bt the formula:

$W_{lost} = T_{surr} *S$

where ;

$T_{surr}$ is the temperature of the surrounding. = 20°C = (20+273.15)K = 293.15 K

From above the change in entropy is:

$\delta \ S = C_p \ In (\frac{T+ \delta \ T }{T}) * \beta V \delta P$

$\delta \ S = 4.19* \ In (\frac{288.15+0.093 }{288.15}) - 1.5*10^{-4} * 1.001 (5-1)* (\frac{1}{9.87})$

$\delta \ S =1.408*10^{-3} \ J/gm.K$

$W_{lost} = T_{surr} *S$

$W_{lost} = 293.15* 1.408*10^{-3} \ J/gm.K$

$W_{lost} = 0.413 \ kJ/kg$

Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

6 0

4 years ago

A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled

Vikentia [17]

Answer:

Force per unit plate area is 0.1344 $N/m^{2}$

Solution:

As per the question:

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Suppose the drag force that exist between each wall and plate is F and F' respectively:

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$\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}$

$\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}$

8 0

3 years ago