Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
#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:
I = 16 kg*m²
Explanation:
Newton's second law for rotation
τ = I * α Formula (1)
where:
τ : It is the moment applied to the body. (Nxm)
I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)
α : It is angular acceleration. (rad/s²)
Kinematics of the wheel
Equation of circular motion uniformly accelerated :
ωf = ω₀+ α*t Formula (2)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (rad)
Data
ω₀ = 0
ωf = 1.2 rad/s
t = 2 s
Angular acceleration of the wheel
We replace data in the formula (2):
ωf = ω₀+ α*t
1.2= 0+ α*(2)
α*(2) = 1.2
α = 1.2 / 2
α = 0.6 rad/s²
Magnitude of the net torque (τ )
τ = F *R
Where:
F = tangential force (N)
R = radio (m)
τ = 80 N *0.12 m
τ = 9.6 N *m
Rotational inertia of the wheel
We replace data in the formula (1):
τ = I * α
9.6 = I *(0.6
)
I = 9.6 / (0.6
)
I = 16 kg*m²
Answer:
The plate's surface charge density is 
Explanation:
Given that,
Speed = 9800 km/s
Distance d= 75 cm
Distance d' =15 cm
Suppose we determine the plate's surface charge density?
We need to calculate the surface charge density
Using work energy theorem


Here, final velocity is zero
...(I)
We know that,


...(II)
From equation (I) and (II)

Charge is negative for electron

Put the value into the formula


Hence, The plate's surface charge density is 
Answer: a) 8.9km/s b) 4.05 x 10^-25 kg
Explanation:
a)
To find the velocity,
V = E/B plug in the numbers we get
V = 1.6kV/m / 0.18 T = 8.9 km/s
b)
To find the ion mass,
m = erB^2/E = 0.125 m x 1.6x10^-19 x (0.18)^2 / ( 1.6 x 10 ^ 3 V/m ) = 4.05 x 10 ^ -25 kg