Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
.................2
put here value and we get
<h3>v = 0.155 m </h3>
so
Magnification will be here as
m =
m =
<h3>m = 1</h3>
Answer:
B
Explanation:
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Answer:
Explanation:
In an electric field E force on charge q
F = Eq , acceleration a = Eq / m
a = 664 x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷
= 636.16 x 10⁸ m /s²
b )
initial velocity u = 0
final velocity v = 1.46 x 10⁶ m/s
v = u + at
1.46 x 10⁶ = 0 + 636.16 x 10⁸ x t
t = 2.29 x 10⁻⁵ s
c )
s = ut + 1/2 a t²
= 0 + .5 x 636.16 x 10⁸ x ( 2.29 x 10⁻⁵ )²
= 1668 x 10⁻²
= 16.68 m
d )
Kinetic energy = 1/2 m v²
= .5 x 1.67 x 10⁻²⁷ x ( 1.46 x 10⁶ )²
= 1.78 x 10⁻¹⁵ J .
Answer:
Using the log combination rules to reduce the famous Sakur-Tetrode equation, The change in entropy is given as:
∆S = NK*ln(V"V$/V").
Where V"V$ is final Volume (Vf) after constraint's removal,
V" is Initial Volume (Vi) before constraint's removal.
Temperature (T) is constant, Internal Energy, U is constant, N and K have their usual notations
Explanation:
Given in the question, the container is an adiabatic container.
For an adiabatic contain, it does not permit heat to the environment due to its stiff walls. This implies that the Internal Energy, U is kept constant(Q = U). The temperature is also constant (Isothermal). Thus, the famous Sakur-Tetrode equation will reduce to ∆S = NK* In(Vf/Vi).
Vf is the volume after the constraint is removed(Vf = V"V$). Vi is the volume occupied before the constraint is removed (Vi = V")