- The net force is greatest at the position of maximum displacement
- The net force is zero when at the equilibrium position
Explanation:
The motion of a spring is a Simple Harmonic Motion, in which the displacement of the end of the spring is given by a periodic function of the form
where A is the amplitude (the maximum displacement), and 
the angular frequency of the motion.
We can analyze the net force acting on the spring by looking at Hooke's law:
where
F is the net force
k is the spring constant
x is the displacement
From the equation, we notice immediately that:
- The net force is the greatest when the displacement x is the greates, so at the position in which the spring has maximum compression or stretching
- The net force is zero when the displacement x is zero, so when the spring crosses the equilibrium position
Learn more about forces:
brainly.com/question/8459017
brainly.com/question/11292757
brainly.com/question/12978926
#LearnwithBrainly
The De Broglie's wavelength of a particle is given by:
where
is the Planck constant
p is the momentum of the particle
In this problem, the momentum of the electron is equal to the product between its mass and its speed:
and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:
So, the answer is True.
The answer is letter C. <span>A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of negative acceleration.
In physics, acceleration can be described as an objects change of velocity. When an object gains velocity, it is positive acceleration, and negative acceleration for the opposite.
</span>
Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.
<span> </span>
Answer:
Acceleration = 5.77 m/s²
Distance cover in 13 seconds = 487.56 meter
Explanation:
Given:
Final velocity of mobile device = 75 m/s
initial velocity of mobile device = 0 m/s
Time taken = 13 seconds
Find:
Acceleration
Distance cover in 13 seconds
Computation:
v = u + at
75 = 0 + (a)(13)
13a = 75
a = 5.77
Acceleration = 5.77 m/s²
s = ut + (1/2)(a)(t²)
s = (0)(t) + (1/2)(5.77)(13²)
Distance cover in 13 seconds = 487.56 meter
Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
R= 12 ohm
2) circuit Q
R = 2.3 ohm
