Answer:
3. Step 1; An action potential depolarizes the axon terminal at the presynaptic membrane
2. Step 2; Calcium ions enter the axon terminal
4. Step 3; Acetylcholine is released from storage vesicles by exocytosis
5. Step 4; Acetylcholine binds to receptors on the postsynaptic membrane
1. Step 5; Chemically gated ion channels on the postsynaptic membrane are opened
Explanation:
3. The cholinergic synapse starts at the point of arrival of an electrochemical impulse or action potentials at the synaptic knob of the axon terminal of a presynaptic neuron membrane
2. The arrival of the action potential at the axon terminal causes the calcium ion Ca²⁺ channels to open and Ca²⁺ enters into the synaptic knob, resulting in the fusion of the presynaptic membrane and synaptic vesicles
4. The fusion enables the release into the synaptic cleft of many acetylcholine (ACh) transmitter molecules by exocytosis
5. Some of the ACh are transported across the synaptic cleft and bind to postsynaptic neuron membrane embedded ACh receptors
1. The binding of the ACh neurotransmitter molecules to receptors on the membrane of the dendrites of a neuron it leads to the opening of ion channels
Poder = (resistencia) x (corrente)²
Poder = (10 ohms) x (5 A)²
<em>Poder = 250 watts </em>(250 Joule por segundo)
2 horas = 7,200 segundos
Energia = (250 joule/seg) x (7,200 seg)
<em>Energia = 1,800,000 Joules</em>
Answer:
A. 91 meters north
Explanation:
Take +y to be north.
Given:
v₀ = 13 m/s
a = 0 m/s²
t = 7 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13 m/s) (7 s) + ½ (0 m/s²) (7 s)²
Δy = 91 m
The displacement is 91 m north.
Answer: The correct answer is "trumpet".
Explanation:
Resonance occurs when the applied frequency on the object is equal to the natural frequency of the object.
Wind instruments use resonance in the air column to amplify the sound. It makes the sound louder. Sound is produced by using lips or vibrating reeds or buzzing sound into the mouthpiece.
Therefore, resonance in air column is used in trumpet musical instruments.
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
