The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole. So yes true
Rhythms that occur faster and slower than the beat are b.<span>not synchronized with the time signature. The synchronization follows the same beat or rhythm. If the time signature say is lower than the original, then the rhythm should be faster. Otherwise, the rhythm is slower than the original one.</span>
Answer:
Autotrophs
Explanation:
When you go down a food chain continuing to ask "what does it eat?" the last living thing that you will land upon is an autotroph.
Autotrophs are the primary producers as they (photoautotrophs) use the energy either from the sun to prepare there food by the process of photosynthesis or, more rarely, obtain chemical energy through oxidation (chemoautotrophs) to make organic substances from inorganic ones.
Autotrophs get consumed by the primary consumers in the food chain.
Answer:
The moon keeps the same face pointing towards the Earth because its rate of spin is tidally locked so that it is synchronized with its rate of revolution (the time needed to complete one orbit). In other words, the moon rotates exactly once every time it circles the Earth.
Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg
