A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
2 answers:
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#1
Volume of lead = 100 cm^3
density of lead = 11.34 g/cm^3
mass of the lead piece = density * volume
so its weight in air will be given as
now the buoyant force on the lead is given by
now as we know that
so by solving it we got
V = 11.22 cm^3
(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N
(iii) Buoyant force = 0.11 N
(iv)since the density of lead block is more than density of water so it will sink inside the water
#2
buoyant force on the lead block is balancing the weight of it
(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N
(iii) Buoyant force = 11.11 N
(iv) since the density of lead is less than the density of mercury so it will float inside mercury
#3
Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid
Look at the picture for the answer
Answer:
λ = 2042 nm
Explanation:
given data
screen distance d = 11 m
spot s = 4.5 cm = 4.5 × m
separation L = 0.5 mm = 0.5 × m
to find out
what is λ
solution
we will find first angle between first max and central bright
that is tan θ = s/d
tan θ = 4.5 × / 11
θ = 0.234
and we know diffraction grating for max
L sinθ = mλ
here we know m = 1 so put all value and find λ
L sinθ = mλ
0.5 × sin(0.234) = 1 λ
λ = 2042.02 × m
λ = 2042 nm