Answer:
The electrons in oxygen are paired while in nitrogen, they are not.
Explanation:
To analyse this we start with writing out the ground state electronic configurations for both elements.
Oxygen: 1s²2s²2p4 meaning the p subshell has the following arrangement of electrons ↑↓ ↑ ↑
Nitrogen : 1s²2s²2p³ meaning the p subshell has the following arrangement of electrons ↑ ↑ ↑
Clearly the paired electron in oxygen will be experiencing repulsion from the electron it shares an orbital with causing it to be removed easily. The electrons in nitrogen are unpaired, each orbital is singly occupied
The force of gravity on the object is 14.47 N
Explanation:
The weight of an object (which is the force of gravity experienced by an object) at a certain location is given by

where
m is the mass of the object
g is the acceleration of gravity at the location of the object
IN this problem, we have:
m = 24.52 kg (mass of the object)
(acceleration of gravity on Pluto)
Substituting, we find the force of gravity on the object:

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Molecules and polyatomic ions are formed by covalent bonds.
Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m
Answer:
7500 Newtons
Explanation:
Mass of the sportscar= 1500 kg
Acceleration of the sportscar= 5m/s^2
Hence, let the Force acting on it be F
