Mechanical energy E = mgh + 1/2mv²
When he starts, let h = 0 ⇒ E₁ = 1/2mv₁²
When he reaches height h ⇒ E₂ = mgh + 1/2mv₂²
Without friction, energy is conserved at all times.
E₁ = E₂
↓
1/2mv₁² = mgh + 1/2mv₂²
↓
1/2v₁² = gh + 1/2v₂²
↓
gh = 1/2(v₁² - v₂²)
↓
h = (v₁² - v₂²) / (2g)
Answer: False
When the diaphragm
contracts, the muscles will also contract and pull upward and increase the size
of the thoracic cavity thus decreases air pressure inside during inspiration. After
the diaphragm contracts, it goes to relaxation, the muscles will also relaxed. It
gets looser and return to its original position higher up in the chest. This increase
the pressure in the chest, which force the air in the lungs out through the
nose.
Explanation:
A splint is lit and held near the opening of the tube, then the stopper is removed to expose the splint to the gas. If the gas is flammable, the mixture ignites. This test is most commonly used to identify hydrogen, which extinguishes with a distinctive 'squeaky pop' sound.
Its because of the static electricity.As we rub, the positive and negative charges are acquired by the plastic rod and duster.
If we bring paper bits near them, the charges possessed by the paper bits get attracted towards the charges in plastic rod.
