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AURORKA [14]
3 years ago
10

Consider a hot-air balloon. The deflated balloon, gondola, and two passengers have a combined mass of 315 kg. When inflated, the

balloon contains 845 m^3 of hot air. Find the temperature (∘C) of the hot air required to lift the balloon off the ground. The air outside the balloon has a temperature of 20.2∘C and a pressure of 0.916×105 Pa. The pressure of the hot air inside the balloon is the same as the pressure of the air outside the balloon. The molar mass of air is 29 g/mol. Ignore the buoyant force on the gondola and the passengers. Hint: First find the density of the air outside the balloon. Then use Archimedes' principle to find the density required for the air inside the balloon. Then find the temperature of the air inside the balloon.
Physics
1 answer:
vladimir2022 [97]3 years ago
3 0

This will help you.

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How do you think car makers can design cars to limit cell phone distractions?
Dafna11 [192]

I have two (2) brilliant ideas:

1). Inside the metal that the body of the car is made of, and also between the two sheets of glass that the windows are made of, install a thin layer of material that absorbs RF (radio-wave) energy . . . like the material in the glass window of your microwave oven.  Then, no radio waves from the cellular base station can get INTO the car, and no radio waves from your phone can get OUT of the car.  The phone can't make a connection to the cellular network, you can't make or receive calls, and you can't connect to Instagram or Brainly, so you might as well just turn it off and save your battery until next time you're outside your car.

2). Somewhere inside the car, like under the dash or in the glove box, install a teeny tiny radio receiver that can recognize the signals coming OUT of your phone.  Connect it to the car's electrical system so that when it hears signals from phones inside the car, it it shuts down the car's motor so you can't start or drive. The car only works when phones inside the car are either turned off or in Airplane Mode.

My ideas are so brilliant that I really should patent them, or copyright them, or whatever you do so that other people have to pay you to use your idea. But if you want to use them, that's OK.  Just go ahead. I won't mind.

8 0
3 years ago
Below is the data from a gas law experiment comparing the pressure and the volume of a gas at a given temperature.
Wewaii [24]

Answer:

The combined gas equation relates three variables pressure, temperature and volume when the number of moles is constant.

The equation is PV / T = constant. Which is valid for a fixed number of moles of the gas.

You can derive the combined gas equation from the combination of Bolye's law, Charles' law and Gay-Lussac's law, which needs some algebra.

Explanation:

9 0
3 years ago
Read 2 more answers
The blades in a blender rotate at a rate of 7700 rpm. when the motor is turned off during operation, the blades slow to rest in
Tpy6a [65]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the beginning)

For this fan, speed at the end = 7700 rpm, speed at the end = 0 .

Change in angular speed = -7700 rpm

Angular acceleration = (-7700 rpm) / (2.5 sec)

<em>Angular acceleration = -3,080 rev per minute / sec</em>

That's a perfectly good and true answer to the question, but the units are ugly.  We really need to fix the units, and convert them into something prettier before we hand in this assignment.

1 rev = 2π radians, and

1 minute = 60 seconds .

So

Angular acceleration =

(-3,080 rev/min-sec) · (2π rad/rev) · (1 min/60 sec)

AngAccel = (-3,080 · 2π · 1 / 60) · (rev·rad·min / min·sec·rev·sec)

AngAccel = ( -102 and 2/3 · π) · (rad/s²)

<em>AngAccel = -322.5 radian/s²</em>

7 0
3 years ago
Where is the magnetic field of a horseshoe magnet the strongest?<br> pls hellp omg
Nuetrik [128]
The answer is slightly left and slightly right of the curved end of the horseshoe.
7 0
3 years ago
Read 2 more answers
Aluminum has a density of
agasfer [191]

<h3><u>Volume is 0.1848 m³</u></h3><h3 />

Explanation:

<h2>Given:</h2>

m = 49.9 kg

ρ = 270 kg/m³

<h2>Required:</h2>

volume

<h2>Equation:</h2>

ρ \:= \:\frac{m}{v}

where: ρ - density

m - mass

v - volume

<h2>Solution:</h2>

Substitute the value of ρ and m

ρ \:= \:\frac{m}{v}

270\: kg/m³\:= \:\frac{49.9\:kg}{v}

(v)\:270\: kg/m³\:= \:49.9\:kg

v\:= \:\frac{49.9\:kg}{270\: kg/m³}

v\:= \:0.1848\:m³

<h2>Final Answer:</h2><h3><u>Volume is 0.1848 m³</u></h3>
3 0
2 years ago
Read 2 more answers
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