Galileo Galilei is one of the key figures in the history of Science, being the first to apply the experimental-mathematical scientific method. He carried out experiments and careful observations in kinematics (his studies on the trajectory of projectiles are famous) and dynamics (it should be noted his careful experiments with inclined planes), establishing the first law of Dynamics (which Newton will later collect and refine in his Principles); and in Astronomy, with which he could unequivocally support the heliocentric theory.
His experiments were addressed by methodologies that allowed him to precisely find his mathematical calculations and to verify theories he was developing over time. His manuscripts were key to disseminate the applied method and extrapolate them to other scientific areas.
Therefore the correct answer is C.
First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
protons
Then, the total charge is the number of protons times the charge of a single proton:
Answer:
3.62m/s and 2.83m/s
Explanation:
Apply conservation of momentum
For vertical component,
Pfy = Piy
m* Vof (sin38) - m*Vgf (sin52) = 0
Divide through by m
Vof(sin38) - Vgf(sin52) = 0
Vof(sin38) = Vgf(sin52)
Vof (sin38/sin52) = Vgf
0.7813Vof = Vgf
For horizontal component
Pxf= Pxi
m* Vof (cos38) - m*Vgf (cos52) = m*4.6
Divide through by m
Vof(cos38) + Vgf(cos52) = 4.6
Recall that
0.7813Vof = Vgf
Vof(cos38) + 0.7813 Vof(cos52) = 4.6
0.7880Vof + 0.4810Vof = 4.
1.269Vof = 4.6
Vof = 4.6/1.269
Vof = 3.62m/s
Recall that
0.7813Vof = Vgf
Vgf = 0.7813 * 3.62
Vgf = 2.83m/s
Answer:
Energy of Photon = 4.091 MeV
Explanation:
From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.
K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)
where,
K.E OF electron = 3.58 MeV
Rest Energy of electron = 0.511 MeV
Rest Energy of positron = 0.511 MeV
K.E OF positron = 3.58 MeV
Energy of Photon = ?
Therefore,
3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)
Energy of Photon = 8.182 MeV/2
<u>Energy of Photon = 4.091 MeV</u>
Answer:
Explanation:
-The chemical formula for Molybdenum (V) Dichromate is 
-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate
-We apply Avogadro's constant to find the number of atoms of oxygen:
Hence, there are 
