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3 years ago

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The basic function of an automobile carburetor is to atomize the gasoline and mix it with air to promote rapid combustion. Assum

e that 25cm 3 of gasoline is atomized into N spherical droplets. Each droplet has a radius of 1.0 × 10 −5 m. Find the total surface area of these N spherical droplets

1 answer:

tatyana61 [14]3 years ago

5 0

Answer:

$A = 7.5 \times 10^6 m^2$

Explanation:

Since volume of all the liquid is always conserved

so here we know that

$V_{total} = N V$

$25 cm^3 = N(\frac{4}{3}\pir^3)$

$25 = N (\frac{4}{3} \pi (1.0 \times 10^{-5})^3)$

$N = 5.97 \times 10^{15}$

now we know that total surface area is given as

$A = N(4\pi r^2)$

$A = (5.97 \times 10^{15})(4\pi (1 \times 10^{-5})^2$

$A = 7.5 \times 10^6 m^2$

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I would pick the only one that makes sense, D. optic fibers

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3 years ago

The Southwest Indian Ridge, shown in red, moves at an average rate of 20 mm/year, making it among the ultraslow spreading ridges

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v = average speed of movement of the Southwest Indian Ridge = 20 mm/year

d = distance moved by the Southwest Indian Ridge = 100 mm

t = number of years required to move distance "d"

distance traveled is given as

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3 years ago

Read 2 more answers

I need answers and solvings to these questions

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1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

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A baseball sits motionless near first base on a baseball diamond. What statement

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Answer:

B. There are no forces acting on the ball.

Explanation:

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Lana wanted to do an investigation to see what kind of birdseed the birds in her neighborhood like best. She purchased three of

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Answer:

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Explanation:

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The independent variable is normally the x-value while the dependent variable is the y-value

In the question, Lana wants to find out the kind of birdseed that the neighborhood birds like (to eat) the most by feeding them different types of birdseed and measuring the level of the birdseed in the feeders at regular intervals

the independent variable is the bird seed types which Lana gives to the birds, to determine the type of bird seeds the birds like

The dependent variable is the level of the seed in the feeder which depends on the type of seeds the birds like.

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