Two students, sitting on frictionless carts, push against each other. Both are initially at rest and the mass of student 1 and t
1 answer:
Answer:
v₂ = v/1.5= 0.667 v
Explanation:
For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.
Initial moment before pushing
p₀ = 0
Final moment after they have been pushed
= m₁ v₁ + m₂ v₂
p₀ =
0 = m₁ v₁ + m₂ v₂
m₁ v₁ = - m₂ v₂
Let's replace
M (-v) = -1.5M v₂
v₂ = v / 1.5
v₂ = 0.667 v
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Answer:
See attached pictures.
Explanation:
See attachments for explanation.
Answer:
A). σ = 3.823 x
/N-
B). C/
C). J
Explanation:
A). We know magnitude of charge per unit area for a conducting plate is given by
where, E is resultant electric field = 1.2 x V/m
is permittivity of free space = 8.85 x
/N-
k is dielectric constant = 3.6
∴
= 3.6 x 8.85 x x 1.2 x
= 3.823 x
/N-
B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by
C/
C).
Area of the plate, A = 2.5
= 2.5 x
diameter of the plate, d = 1.8 mm
= 1800 m
∴ Total energy stored in the capacitor
J
Answer:
1.36
Explanation:
= Index of refraction of air = 1
= Index of refraction of plastic = ?
i = angle of incidence in air = 32.0° deg
r = angle of refraction in plastic = 23.0° deg
Using Snell's law
Sini =
Sinr
(1) SIn32.0 = Sin23.0
= 1.36