Answer:
Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s
Explanation:
Given;
initial speed of proton, u = 2.5 x 10⁵ m/s
initial potential, V = 1500 V
mass of proton = 1.67 x 10⁻²⁷ kg
Work done, W = eV= ΔK.E = ¹/₂mu²
eV = ¹/₂mu² (J)
where;
e is the charge of the proton in coulombs
V is the electric potential in volts
m is the mass of the proton in kg
u is the speed of the proton in m/s
Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s
Answer:
number of moles = 27.34 moles
the temperature of gas after it undergoes the isobaric expansion = 605 K
Explanation:
Given that:
V = 0.25 m³
P = 250 kPa
T = 275 K
V₂ = 0.55 m³
P₂ = 760 kPa
a)
Using ideal gas equation ; PV = nRT
b) To calculate the temperature of gas after it undergoes the isobaric expansion; we have:
creo que soy yo pero lo veo en ingles
Between the stars' absolute magnitudes<span> or </span>luminosities<span> versus their </span>stellar classifications<span> or </span>effective temperatures<span>. </span>
Magnetic force and Gravity are non-contact forces among friction, electrostatic force, magnetic force and gravity.
