The electron is accelerated through a potential difference of
, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
is the potential difference
Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
Answer:
x = 1.6 + 1.7 t^2 omitting signs
a) at t = 0 x = 1.6 m
b) V = d x / d t = 3.4 t
at t = 0 V = 0
c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)
d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)
Assuming the gas behaves ideally,
PV/T = constant. P will also be constant in this giving us:
V₁/T₁ = V₂/T₂
40/320 = 20/T₂
T₂ = 160 K
The answer is A.
Answer:
R=m*g-∀fl*g*l3
Explanation:
<em>An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which reads W as the weight. The top of the block is a height h below the surface of the fluid. The correct equation for the reading of the scale is</em>
From Archimedes' principle we know that a body when immersed in a fluid, fully or partially, experiences an the upward buoyant force equal to the weight of the fluid displaced. As the body is fully submerged in water, volume of water displaced
density of iron =mass/ volume
rho=m/l3
mass=rhol3
weight fluid=rhofluid*g*Volume
weight of fluid=rhofluid*g*l3
F=∀fl*g*l3
Downward force is weight of iron
w=m*g
Reading on the spring scale
R=w-F
R=m*g-∀fl*g*l3
m=mass of iron
g=acceleration due to ravity
rhfld=density of fluid
l3=volume of fluid displaced
Complete Question:
Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:
A. 2F
B. F/4
C. F/2
D. F
E. 4F
Answer:
D.
Explanation:
If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):
As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.
As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.
