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Zepler [3.9K]
3 years ago
13

A hot air balloon has a mass of 300 k gincluding the basket. The hot air inside the balloon weighs 17,000 N. Determine the maxim

um load that the balloon is able to lift displaces 25,000 N of cold air. Use 10 m/s​2 for gravity.
Physics
1 answer:
padilas [110]3 years ago
7 0

Answer:

500 kg

Explanation:

If 17000 N hot air weight displaces 25000 N cold air weight, then the difference of 25000 - 17000 = 8000 N in weight would result in the buoyant force that lift the whole system.

If g = 10m/s2 then the total mass that it can lift (including the 300kg basket) is

8000 / 10 = 800 kg

So the maximum additional load that the balloon is able to lift is

800 - 300 = 500 kg

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3 years ago
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A balloon is rubbed against a sweater. Which of the following describes the result of this interaction?(1 point)
Ghella [55]

Fibers of the sweater lose electrons because electrons are leave it.

One of the ways of charging a body is by friction. Charges are transferred from one body to another when an object is rubbed against another. This is charging by friction.

A sweater has negative charges hence when the balloon is rubbed against the sweater, fibers of the sweater lose electrons because electrons are leave it.

Learn more: brainly.com/question/830809

4 0
2 years ago
An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e
timama [110]

Answer:

K = 80.75 MeV    

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}

<em>where E_{ph}: is the photon energy, E_{0p} and E_{0ap}: are the rest energies of the proton and the antiproton, respectively, equals to m₀c², K_{p} and K_{ap}: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>        

Therefore the kinetic energy of the antiproton is:    

K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}

<u>The proton mass is equal to the antiproton mass, so</u>:

K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}  

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV

K_{ap} = 80.75 MeV              

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

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Komok [63]

Answer:

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I took the test and got this correct. Hopefully this helps you!

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