Answer:
x=22.33m
Explanation:
Kinematics equation for constant deceleration:

Atoms like carbon and nitrogen do not form ions because the electronegativity of these atoms are not that high nor very low which means electrons are fairly stable in the atom. While chlorine has very high electronegativity and for sodium very low, atoms tend to receive or release electrons.
When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>. (choice-D)
The increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.
<h3>What is gravitational potential energy?</h3>
The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.
The potential energy increases by 3773 J
PE₂-PE₁=mg(h₂-h₁)
3773 J = 35.0 × 9.81 × (h₂-h₁)
(h₂-h₁) = 10.98
Case 2 ;
ΔPE =?
ΔPE=mg(h₂-h₁)
ΔPE=56.0 × 9.81 ×10.98
ΔPE=6031.97 J.
Hence, the increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.
To learn more about the gravitational potential energy, refer;
brainly.com/question/3884855#SPJ1
#SPJ1
All of the following
involve waves of electromagnetic energy except the rumble of thunder during a storm.
Electromagnetic waves<span> <span>are
used to transmit long/short/FM wavelength radio </span>waves, and TV/telephone/wireless signals or energies. They are
also responsible for transmiting energy in the form of microwaves, infrared radiation<span> (IR), visible light (VIS),
ultraviolet light (UV), X-rays, and gamma rays.</span></span>
The correct answer between all
the choices given is the second choice or letter B. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.