#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
Answer:
Explanation:
From Newton's second law,
where 
is the force, 
is the mass and 
is the acceleration.
From Hooke's law,
where 
is the spring constant and 
is the displacement function measured from the origin. The negative sign indicates the force acts in opposite direction to the displacement. In fact, it is a restoring force; it acts to return the spring to its original undisturbed position.
Since both forces are the same,
The implication of this is that the acceleration is proportional to the displacement but opposite to it. That last statement is the definition of a simple harmonic motion which this is.
The ratio 
is a constant except in situations where the mass is varying (say, the mass on the spring is a decaying material).
Answer:
The final size is approximately equal to the initial size due to a very small relative increase of 
in its size
Solution:
As per the question:
The energy of the proton beam, E = 250 GeV =
Distance covered by photon, d = 1 km = 1000 m
Mass of proton, 
The initial size of the wave packet, 
Now,
This is relativistic in nature
The rest mass energy associated with the proton is given by:
This energy of proton is 
Thus the speed of the proton, v
Now, the time taken to cover 1 km = 1000 m of the distance:
T = 
T = 
Now, in accordance to the dispersion factor;
Thus the increase in wave packet's width is relatively quite small.
Hence, we can say that:
where
= final width
Answer:
a. 8.33 x 10 ⁻⁶ Pa
b. 8.19 x 10 ⁻¹¹ atm
c. 1.65 x 10 ⁻¹⁰ atm
d. 2.778 x 10 ⁻¹⁴ kg / m²
Explanation:
Given:
a.
I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s
P rad = I / us
P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s
P rad = 8.33 x 10 ⁻⁶ Pa
b.
P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]
P rad = 8.19 x 10 ⁻¹¹ atm
c.
P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]
P rad = 1.67 x 10 ⁻⁵ Pa
P₁ = 1.013 x 10 ⁵ Pa /atm
P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm
d.
P rad = I / us
ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²
ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²
