Express the volume expansivity and the isothermal compressibility as functions of density rho and its partial derivatives. For w
1 answer:
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Answer:
speed of the proton is 6.286 × m/s
Explanation:
given data
charge q= −60.0 nC
inner radius a = 20.0 cm
outer radius b = 24.0 cm
charge density ρ = −2.05 µC/m³
to find out
What is the speed of the proton
solution
we know that force on the proton due to this electric field is express as
F = q × E ...................1
here F is force and q is charge and E is electric filed so
if v be the speed of the proton in circular orbit than force will be
F = ....................2
from equation 1 and 2
q × E = .......................3
so
here total charge Q on shell is
Q = ρ × V
here ρ is density and V is volume
Q =
put here value
Q =
Q = 50.01 × C
and
total charge enclosed by Gaussian surface is
qin = q + Q
qin = −60 × C - 50.01 ×
C
qin = - 110.01 × C
and
from Gauss law
E ×4×π×b² =
E =
E =
E = 17189.06 N/C
so
from equation 3
q × E =
v =
v =
v = 6.286 × m/s
so speed of the proton is 6.286 × m/s
Answer:
0.064 m
Explanation:
When a spring is stretched/compressed, the force exerted in the spring is related to the elongation of the spring by the equation:
where:
F is the force applied
k is the spring constant
x is the stretching/compression of the spring
In this problem, we have:
is the force applied by the kid on the spring
k = 128 N/m is the spring constant
Solving for x, we find how far the spring stretches: