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:
E =230.4 MJ
Explanation:
As 1 mole of electron = 6X 10^23 particles.
charge of an electron is 1.6 X 10 ^-19 C
Finding Charge:
(6X10^23 ) (2.7)(1.6X10^-19 C)
i.e. 192 K C
now to find the energy released from electrons
V=E/q
E=V X q
i.e E = 120 V X 192 K C
E =230.4 MJ
A. 20m/s because the unit for velocity is m/s
Answer:
(i) 
(ii) 
Explanation:
Let t be the average thickness of the sheet.
Given that:
Density of the aluminum sheet is 
Mass of sheet = 60.7 g
Length of sheet = 50.0 cm
Width of sheet = 30.0 cm
(i) Using, Density=Mass/Volume


Hence, the volume of the sheet is
.
(ii) Now, as this aluminum sheet is in the shape of a cuboid, so the volume of the sheet is



Hence, the average thickness of the sheet is
.