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<u>Answer: Koala bears are considered herbivores, or as in the scientific name, arboreal herbivorous marsupial, marsupial because it also carries it's babies around in a pouch. Koala bears are also native to Australia, which eucalyptus leaves are also native to.
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Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
The atmospheric P is greater than the P in the flask, since
the Hg level is lacking down lower on the side open to the atmosphere.
43.4 cm x (10 mm / 1 cm) = 435 mm
the density of Hg is 13.6 / 0.791 = 17.2 times better than the liquid in the
manometer. This means that 1 mmHg = 17.2 mm of manometer liquid.
435 mm manometer liquid x (1 mm Hg / 17.2 mm manometer liquid) = 25.3 mm
Hg
The pressure in the flask is 755 - 25.3 = 729.7 mmHg.
729.7 mmHg x (1 atm / 760 mmHg ) = 0.960 atm.
Answer:
7500 m/s
Explanation:
Centripetal acceleration = gravity
v² / r = GM / r²
v = √(GM / r)
Given:
G = 6.67×10⁻¹¹ m³/kg/s²
M = 5.98×10²⁴ kg
r = 6.8×10⁵ + 6.357×10⁶ = 7.037×10⁶ m
v = √(6.67×10⁻¹¹ (5.98×10²⁴) / (7.037×10⁶))
v = 7500
The orbital velocity is 7500 m/s.
KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required
to maintain a specified surface temperature for water and air flows.
FIND: Convection coefficients for the water and air flow convection processes, hw and ha,
respectively.
ASSUMPTIONS: Flow is cross-wise over cylinder which is very long in the direction
normal to flow.
The convection heat rate from the cylinder per unit length of the cylinder has
the form
q' = h*(pi*D)*(Ts-Tinf)
and solving for the heat transfer convection coefficient, find
Water
hw = q'/((pi*D)*(Ts-Tinf))
hw = (38*10^3 W/m) / ((pi*(0.030m))*(80-25)C)=
7330.77314 W/m^2K
Air
ha = (400W/m) / ((pi*(0.030m))*(80-25)C)=<span>
77.166033 </span> W/m^2K
COMMENTS: Note that the air velocity is 10 times that of the water flow, yet
hw ≈ 95 × ha.
These values for the convection coefficient are typical for forced convection heat transfer with
liquids and gases
Watter is a better convective heat transfer media than air
