If an object is thrown in an upward direction from the top of a building 1.6 x 10^2 ft. high at an initial velocity of 21.82 mi/
2 answers:
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(a) 198 g
When the rock is submerged into the water, there are two forces acting on the rock:
- its weight, equal to (m=mass, g=acceleration of gravity), downward
- the buoyant force, equal to (
=mass of water displaced), upward
So the resultant force, which is the apparent weight of the rock (W'), is
which can be rewritten as
where m' is the apparent mass of the rock. Using:
m = 540 g
m' = 342 g
we find the mass of water displaced
(b)
If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.
The volume of water displaced is given by
where
is the mass of the water displaced
is the density of the water
Substituting,
And so this is also the volume of the rock.
(c)
The average density of the rock is given by
where
m = 540 g = 0.540 kg is the mass of the rock
is its volume
Substituting into the equation, we find
Answer:
Explanation:
= Velocity of one lump =
= Velocity of the other lump =
m = Mass of each lump =
The collision is perfectly inelastic as the lumps stick to each other so we have the relation
The velocity of the stuck-together lump just after the collision is .
Answer:
a) the magnitude of the force is
F= Q() and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E =
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = , where p = q × s
E(r) = where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²