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Svetllana [295]

3 years ago

8

Convert to the fractional equivalent and reduce 21.12

2 answers:

dem82 [27]3 years ago

7 0

Answer:

528/25

Explanation:

21.12 equals to 2112/100 if we divide twice 2112 and 100 in 2 we have: 2112/100=1056/50=528/25

nignag [31]3 years ago

3 0

The decimal point is placed after two digits starting from the end. For each decimal place, we can write the number divided by 100.

21.12 can be written as $\frac{2112}{100}$ .

Divide the numerator and denominator by 2:

$\frac{2112}{100}= \frac{156}{50}$

The numerator and denominator can be divided by 2 again:

$\frac{78}{25}$

There is no other common factor between numerator and denominator other than 1. Hence, it is the reduced form.

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The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivi

ahrayia [7]

To calculate and solve the problem it is necessary to apply the concepts related to resistance and resistivity.

The equation that is responsible for relating the two variables is:

$R = \rho \frac{L}{A}$

Where,

R= Resistance of the conductor

$\rho =$ Resistivity of the conductor material

L = Length

A = Cross-sectional area of conductor

With the previous values the area of the muscle (Real Muscle-82%)is,

$A_m = (0.82)\pi r^2 = (0.82)\pi (12/2*10^{-2})^2$

$A_m = 9.274*10^{-3}m^2$

Using the equation from Resistance we have that at the muscle the value is:

$R_m = \rho \frac{L}{A}$

$R_m = \frac{13*(0.4)}{9.274*10^{-3}}$

$R_m = 560.70\Omega$

At the same time we can make the same process to calculate the resistance of the fat, then

$A_m = (0.18)\pi r^2 = (0.18)\pi (12/2*10^{-2})^2$

$A_m = 2.0357*10^{-3}m^2$

The resistance of the fat would be,

$R_f = \rho \frac{L}{A}$

$R_f = \frac{25*(0.4)}{2.0357*10^{-3}}$

$R_f = 4912.3\Omega$

Then the total resistance in a set as the previously writen, i.e, in parallel is:

$R=\frac{R_mR_f}{R_m+R_f}$

$R = \frac{(560.70)(4912.3)}{4912.3+560.70}$

$R = 502.62\Omega$

We can here apply Ohm's law, then

$I= \frac{V}{R}$

$I = \frac{1.5}{502.62}$

$I = 2.984*10^{-3}A$

$I = 2.984mA$

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Explanation:

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A jet aircraft is traveling at 262 m/s in horizontal flight. The engine takes in air at a rate of 85.9 kg/s and burns fuel at a

Fittoniya [83]

Answer:

$F_T=60132.52N$

$P=15814852.76W$

Explanation:

From the question we are told that

Velocity of aircraft $V=263m/s$

Engine air intake rate $\triangle M_a=85.9kg/s$

Fuel burn rate $\triangle M_f =3.92kg/s$

Velocity of exhaust gas $V_e =921m/s$

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$F_T=(89.82 *921)-(85.9)*(263)$

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Generally the Rocket's delivered power is mathematically given by

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$P=V*F_T$

$P=263*60132.52N$

$P=15814852.76W.$

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Answer:

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