Newton’s Second Law concerns the generation of force based on an object’s mass and acceleration, as described by the equation F=ma.
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C₂H₃O₂⁻ is an anion.
<u>Explanation:</u>
NaC₂H₃O₂(s) → Na⁺(aq) + C₂H₃O₂⁻(aq)
NaC₂H₃O₂ when dissociated, yields Na⁺ and C₂H₃O₂⁻.
Anion is a negatively charged ion.
In this case, C₂H₃O₂⁻ is an anion.
Answer:
a) the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis
b) the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)
Explanation:
a) the polarization the determined wave oscillates the electric field, which is the z axis
As the wave travels on the negative x-axis and the magnetic field is perpendicular, this field goes on the positive y-axis
the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis
be) in the case of a polarization in the xi plane the magnetic field must go in the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)
Answer:
<em>The period of the motion will still be equal to T.</em>
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Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as
where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, <em>increasing the amplitude has no effect on the period of the mass and spring system.</em>
Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
