Answer:
The direct answer to the question as written is as follows: nothing happens to gravity when someone jumps up - gravity continues exerting a force on the body of that particular someone proportional to (mass of someone) x (mass of Earth) / (distance squared). What you might be asking, however, is what is the net force acting on the body of someone jumping up. At the moment of someone jumping up there is an upward acceleration, i.e., an upward-directed force which counteracts the gravitational force - this is the net force ( a result of the jump force minus gravity). From that moment on, only gravity acts on the body. The someone moves upward gradually decelerating to the downward gravitational acceleration until they reaches the peak of the jump (zero velocity). Then, back to Earth.
consider the motion along the X-direction
X = horizontal displacement = 80 m
= initial velocity along the x-direction = v Cos60
t = time of travel
using the equation
X = t
80 = (v Cos60) (t)
t = 160/v eq-1
consider the motion in vertical direction :
Y = vertical displacement = 20 m
= initial velocity in Y-direction = v Sin60
a = acceleration = - 9.8 m/s²
t = time of travel = 160/v
using the equation
Y = t + (0.5) a t²
20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²
v = 32.5 m/s
Ionization energy, according to <span>chem.libretexts.org,</span><span> is the quantity of </span>energy<span> that an isolated, gaseous atom in the ground electronic state must absorb to discharge an electron, resulting in a cation. This </span>energy<span> is usually expressed in kJ/mol, or the amount of </span>energy<span> it takes for all the atoms in a mole to lose one electron each.</span>
Sup Milk,
Sublimation = Energy is absorbed and a solid turns to a gas.
Condensation = Energy is released and a gas changes to a liquid.
Evaporation = Energy is absorbed into a liquid to turn it into a gas.
I got -3.6 m/s but I had to do conservation of momentum for this question. Which involves Newtons third law but with simply that law I do not know how to complete this question. If you would like me to post my work I will though! Sorry
