(i) Mass of box = 2.3 kg Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg. Total mass = 2.3 + 0.04032 = 2.34032 kg In correct significant figure mass = 2.3 kg (as least decimal) (ii) Difference in mass of gold pieces = 0.02 g In correct significant figure (2 significant fig. minimum decimaRead more
(i) Mass of box = 2.3 kg
Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg.
Total mass = 2.3 + 0.04032 = 2.34032 kg
In correct significant figure mass = 2.3 kg (as least decimal)
(ii) Difference in mass of gold pieces = 0.02 g
In correct significant figure (2 significant fig. minimum decimal) will be 0.02 g.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces 20.15 g and 20.17 g are added to the box. (i) What is the total mass of the box ? (ii) The difference in masses of the pieces to correct significant figures.
(i) Mass of box = 2.3 kg Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg. Total mass = 2.3 + 0.04032 = 2.34032 kg In correct significant figure mass = 2.3 kg (as least decimal) (ii) Difference in mass of gold pieces = 0.02 g In correct significant figure (2 significant fig. minimum decimaRead more
(i) Mass of box = 2.3 kg
Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg.
Total mass = 2.3 + 0.04032 = 2.34032 kg
In correct significant figure mass = 2.3 kg (as least decimal)
(ii) Difference in mass of gold pieces = 0.02 g
See lessIn correct significant figure (2 significant fig. minimum decimal) will be 0.02 g.
The sides of a rectangle are (10..5 ± 0.2) cm and (5.2 ± 0.1) cm. Calculate its perimeter with error limits.
P = 2 (l + b) ± 2 (∆l + ∆b) = 2(10.5 + 5.2) ± 2(0.2 + 0.1) = (31.4 ± 0.6) cm.
P = 2 (l + b) ± 2 (∆l + ∆b)
See less= 2(10.5 + 5.2) ± 2(0.2 + 0.1)
= (31.4 ± 0.6) cm.
Determine the number of light years in one metre.
1 Iy = 9.46 × 10¹⁵ m 1 m = 1/9.46x10¹⁵ = 1.057 x 10⁻16 Iy
1 Iy = 9.46 × 10¹⁵ m
See less1 m = 1/9.46×10¹⁵ = 1.057 x 10⁻16 Iy
Magnitude of force experienced by an object moving with speed v is given by F = kv². Find dimensions of k.
[k] = [F]/[v²] = M¹L¹T⁻²/ M⁰L²T⁻² = [M¹ L⁻¹]
[k] = [F]/[v²] = M¹L¹T⁻²/ M⁰L²T⁻² = [M¹ L⁻¹]
See lessIn Vander Wall’s gas equation (P+a/v²) (v-b) = RT. Determine the dimensions of a and b.
Since dimensionally similar quantities can only be added [P] = [a/v²] ⇒[a]=[PV²]=[M¹L⁵T²] [b] = [v] = [L³]
Since dimensionally similar quantities can only be added
See less[P] = [a/v²] ⇒[a]=[PV²]=[M¹L⁵T²]
[b] = [v] = [L³]