The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F
<h3>How to determine the coefficient of linear expansion</h3>
From the question given above, the following data were obtained
- Original diameter (L₁) = 10 m
- Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
- Change in temperature (∆T) = 90 °F
- Coefficient of linear expansion (α) =?
The coefficient of linear expansion can be obtained as illustrated below:
α = ∆L / L₁∆T
α = 0.015 / (10 × 90)
α = 0.015 / 900
α = 1.67×10¯⁵ /°F
Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F
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Answer:
a) S = 2.35 10³ J/m²2
,
b)and the tape recorder must be in the positive Z-axis direction.
the answer is 5
c) the direction of the positive x axis
Explanation:
a) The Poynting vector or intensity of an electromagnetic wave is
S = 1 /μ₀ E x B
if we use that the fields are in phase
B = E / c
we substitute
S = E² /μ₀ c
let's calculate
s = 941 2 / (4π 10⁻⁷ 3 10⁸)
S = 2.35 10³ J/m²2
b) the two fields are perpendicular to each other and in the direction of propagation of the radiation
In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.
the answer is 5
C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis
Answer:
340.67 kgm²/s
Explanation:
R = Radius of merry-go-round = 1.9 m
I = Moment of inertia = 209 kgm²
= Initial angular velocity = 1.63 rad/s
m = Mass of person = 73 kg
v = Velocity = 4.8 m/s
Initial angular momentum is given by
The initial angular momentum of the merry-go-round is 340.67 kgm²/s
Colder in Alaska, warmer in Mexico.
Answer:
In 0.5 seconds.
Explanation:
The time would be the same because it only depends on the height and the vertical component of the initial velocity. This is of course because each direction must be treated independently. Since between both cases only the horizontal speed changes, the height is the same and the vertical component of the initial velocity is null for both, the time to fall is the same.
