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Igoryamba
3 years ago
13

What forces were exerted during the air resistance experiment?

Physics
1 answer:
zepelin [54]3 years ago
6 0

The forces that were exerted during the air resistance was the gravity.

<u>Explanation:</u>

Air resistance is the frictional power air applies against a moving article. As an article moves, air opposition backs it off. The quicker the article's movement, the more prominent the air obstruction applied against it.

The force acting upon the objects dropped in the air was air resistance and gravity. The force acting upon the objects dropped in the vacuum was gravity only.

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At what point is the KINETIC ENERGY at its highest?
igomit [66]

Answer:

An object has the MOST kinetic energy when it's movement is the GREATEST.

Explanation:

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2 years ago
A 783kg elevator rises straight up 164 meters. What is the potential energy of the elevator?
Vilka [71]

Answer:

potential \: energy = mgh \\ m = 783 \\ g = 10 \\ h = 164 \\ pe = 783 \times 10 \times 164 \\ =  7830 \times 164 \\  = 1284120 \: joule \\ thank \: you

8 0
2 years ago
Read 2 more answers
A motorcycle running on gasoline wastes a large amount of energy mainly as A) heat energy and sound energy. B) light energy and
vlada-n [284]

A motorcycle mainly wastes energy as heat <u>energy</u> and <u>sound</u> energy. In the engine, chemical energy is transformed into mechanical energy. However, the engine is inefficient and much of the chemical energy is lost as heat energy. Also, some of the energy is transformed to sound energy. This explains why the motorcycle is noisy and has an exhaust pipe.

3 0
3 years ago
A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25
Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress  = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter  d₂ =12in  d₁= 12 -4in = 8in

4 -in.-thick  

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

           = 66/ (П* (d₂²-d₁²)/4

           =66/ (3.142* (12²-8²)/4

          = 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

           = 66* cos 25/ (П* (d₂²-d₁²)/4

           =59.82/ (3.142* (12²-8²)/4

          = 59.82/62.84 = 0.952kips/in²

shearing stress  = tangential stress /2

                           = T/2 = 0.952/2 = 0.476 kips/in²

8 0
3 years ago
calculate the mass of displaced water when a piece of 30cm Iceberg with surface area 1000 cm^2 floats on water density of ice is
sweet [91]

Vi = As * h = 1000 * 30 = 30,000 cm^3 = Vol. of the ice.

Vb = (Di/Dw) * Vi = (0.9/1.0) * 30,000 = 27,000 cm^3 = Vol. below surface - Vol. of water displaced.

27,000cm^3 * 1g/cm^3 = 27,000 grams = 27 kg = Mass of water displaced.

3 0
2 years ago
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