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3 years ago

8

g A circular oil slick of uniform thickness is caused by a spill of one cubic meter of oil. The thickness of the oil slick is de

creasing at a rate of 0.1 centimeters per hour. At what rate is the radius of the slick increasing when the radius is 8 meters?

1 answer:

Anika [276]3 years ago

4 0

Answer:

the rate of increase of radius is dR/dt = 0.804 m/hour = 80.4 cm/hour

Explanation:

the slick of oil can be modelled as a cylinder of radius R and thickness h, therefore the volume V is

V = πR² * h

thus

h = V / (πR²)

Considering that the volume of the slick remains constant, the rate of change of radius will be

dh/dt = V d[1/(πR²)]/dt

dh/dt = (V/π) (-2)/R³ *dR/dt

therefore

dR/dt = (-dh/dt)* (R³/2) * (π/V)

where dR/dt = rate of increase of the radius , (-dh/dt)= rate of decrease of thickness

when the radius is R=8 m , dR/dt is

dR/dt = (-dh/dt)* (R³/2) * (π/V) = 0.1 cm/hour *(8m)³/2 * π/1m³ *(1m/100 cm)= 0.804 m/hour = 80.4 cm/hour

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A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

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Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

$F_f= \mu N_2$

where :

$F_f =$ friction force

$\mu$ = coefficient of friction

$N_2$ = normal friction

Given that:

$\mu$ = 0.3

$F_f =$ 0.3 $N_2$

Using the equation of equilibrium along horizontal direction.

$\sum f_x = 0$

P - $F_f =$ 0

P = 0.3 $N_2$ ----- Equation (1)

To determine the moment about point B ; we have the expression

$\sum M_B = 0$

0 = $N_2*70-W*35-P*100$

where;

P = horizontal force

$N_2$ = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = $N_2*70-650*35-100*P$

$P = \frac{70 N_2-22750}{100} ----- equation (2)$

Replacing $\frac{70 N_2-22750}{100}$ for P in equation (1)

$\frac{70N_2 -22750}{100} =0.3 N_2$

$N_2 = \frac{22750}{40}$ $N_2 = 568.75 \ N$

Plugging the value of $N_2 = 568.75 \ N$ in equation (2)

$P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}$

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

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Taking the moments about point A:

$\sum M_A = 0$

-(P × 100)+ (W×35) = 0

P = $\frac{W*35}{100}$

Replacing 650 N for W

$P = \frac{650*35}{100}$

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

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3 years ago

A vertical piston-cylinder assembly contains water. The piston has a mass of 2 kg and a diameter of 7 cm . Assume atmospheric pr

Step2247 [10]

Answer:

for a) F= 744.97 N

for a) F= 167.85 N

for a) F= 764.57 N

Explanation:

the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.

Then from steam tables

T= 120°C → P required=Pr= 198.67 kPa

then the pressure developed by the piston is

P = (m*g + F)/A

where m= mass of the piston ,g= gravity F= force required and A= area of the piston

then

Pr = P = (m*g + F)/A

F = Pr*A-m*g

since A= π/4*D²

F =π/4* Pr*D²-m*g

replacing values

F =π/4* Pr*D²-m*g = π/4*198.67 *10³Pa*(0.07m)² -2kg* 9.8m/s²

F= 744.97 N

b) for T₂=80°C → Pr₂=47.41 kPa

F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²

F₂= 167.85 N

c) for m=0 (mass of the piston neglected) ,the force required is

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F₃ =764.57 N

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3 years ago

A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35

lukranit [14]

Answer:

$P_O = 0.989 watt = 19.9 dBm$

Explanation:

Given data:

P_1 power = 20 dBm = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor $= 10 log \frac{P_1}{P_f}$

solving for final power

$20 = 10 log\frac{P_1}{P_f}$

$2 = log \frac{P_1}{P_f}$

$100 = \frac{0.1}{P_f}$

$P_f = 0.001 watt = 0 dBm$

Directivity $D = 10 \frac{P_f}{P_b}$

$35 = 10 \frac{0.001}{P_b}$

$P_b = 3.162 \times 10^{-7} wattt$

output Power $= P_1 -P_f - P_b$

$= 0.1 - 0.001 - 3.162 \times 10^{-7}$

$P_O = 0.989 watt = 19.9 dBm$

6 0

3 years ago