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3 years ago

Which of the following air conditions would be least likely to have precipitation?

Physics

2 answers:

Lynna [10]3 years ago

7 0

The best and most correct answer among the choices provided by the question is the first choice "warm, dry air"

In meteorology, precipitation<span> is any product of the condensation of atmospheric water vapor that falls under gravity. The main forms of </span>precipitation<span> include drizzle, rain, sleet, snow, graupel and hail.</span>

I hope my answer has come to your help. God bless and have a nice day ahead!

Ghella [55]3 years ago

4 0

Answer: the answer is cold, dry air

Explanation: i took the test and this showed as the correct answer

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a 1.05 kg water bottle is sitting on the teacher's desk it is .20 m tall and has a radius of .03 m find the force that the water

arsen [322]

The force applied would be 1.05*9.8 = 10.3 N
the pressure is equal to F/a
area will be πr^2 = 0.002826
thus pressure will be = 10.3/0.002826= 3644.72 N/m^2

4 0

3 years ago

I need answers and solvings to these questions

den301095 [7]

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:

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)

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

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

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

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

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

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

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

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

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:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago

Multi celled organisms have a ______ surface area to volume ratio than single celled organisms.

True [87]

A. Larger. It is larger Bc they r all larger than the other

6 0

3 years ago

the force between your feet and the floor is greater while standing on your tiptoes than while standing flat on your feet

Elodia [21]

Yes thats correct....becuase all of your weight is concentrated on a small area compared to the larger surface area of your feet!
is that what your question was?

4 0

3 years ago

Read 2 more answers

What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.5 T?

Brilliant_brown [7]

Answer:

The acceleration of the proton is 9.353 x 10⁸ m/s²

Explanation:

Given;

speed of the proton, u = 6.5 m/s

magnetic field strength, B = 1.5 T

The force of the proton is given by;

F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

$a = \frac{qvB}{m}\\\\a = \frac{(1.602*10^{-19})(6.5)(1.5)}{1.67*10^{-27}}\\\\a = 9.353*10^8 \ m/s^2$

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²

4 0

3 years ago