applied forces would be push for example.
normal forces would seem to be a force such as gravity.
friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.
Answer:
a) force between them is attraction, b) F = 1.125 10⁻² N
Explanation:
In this case the electric force is given by Coulomb's law
F =
In the exercise they give us the values of the loads
q1 = - 10 mC = -10 10⁻³ C
q2 = 5 mC = 5 10⁻³ C
d = 20 cm = 0.20 m
let's calculate
F = 9 10⁹ 
F = 1.125 10⁻² N
To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction
Answer:
W = 9.93 10² N
Explanation:
To solve this exercise we must use the concept of density
ρ = m / V
the tabulated density of copper is rho = 8966 kg / m³
let's find the volume of the cylindrical tube
V = A L
V = π (R_ext ² - R_int ²) L
let's calculate
V = π (4² - 2²) 10⁻⁴ 3
V = 1.13 10⁻² m³
m = ρ V
m = 8966 1.13 10⁻²
m = 1.01 10² kg
the weight of the tube
W = mg
W = 1.01 10² 9.8
W = 9.93 10² N
(1) The harmonic number for the mode of oscillation is 3.
(2) The pitch (frequency) of the sound is 579.55 Hz
(3) The level of the water inside the vertical pipe is 0.1 m.
<h3>The harmonic number</h3>
The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.
<h3>Frequency of the wave</h3>
The pitch (frequency) of the sound is calculated from third harmonic formula;
f = 3v/4L
where;
- v is speed of sound
- L is length of the pipe
f = (3 x 340) / (4 x 0.44)
f = 579.55 Hz
<h3>level of the water</h3>
wave equation for first harmonic of a closed pipe is given as
f = v/(4L)
251.1 = 340/(4L)
4L = 340/251.1
4L = 1.35
L = 1.35/4
L = 0.34 m
level of water = 0.44 m - 0.34 m = 0.1 m
Thus, the level of the water inside the vertical pipe is 0.1 m.
Learn more about harmonics of closed pipes here: brainly.com/question/27248821
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The synapse is actually the link between 2 neurons. Now when
an action potential contacts the synaptic knob of a neuron, the voltage-gate
calcium channels are unlocked, resulting in an influx of positively charged
calcium ions into the cell. This makes the vesicles containing
neurotransmitters, for example acetylcholine, to travel towards the
pre-synaptic membrane. When the vesicle arrives at the membrane, the contents
are released into the synaptic cleft by exocytosis. Neurotransmitters disperse
across the space, down to its concentration gradient, up until it reaches the
post-synaptic membrane, where it connects to the correct neuroreceptors. Connecting
to the neuroreceptors results in depolarisation in the post-syanaptic neuron as
voltage-gated sodium channels are also opened, and the positively charged
sodium ions travel into the cell. When adequate neurotransmitters bind to
neuroreceptors, the post-synaptic membrane overcame the threshold level of
depolarisation and an action potential is made and the impulse is transmitted.
