No option to choose from? I'd say history graph
The electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.
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Electric field of the positive particle</h3>
The electric field is calculated as follows;
E = kq/r²
where;
- r is the distance between the charges
- k is Coulomb's constant
- q is magnitude of the charge
midpoint of 3.08 m, x = 1.54 mm
r(1.54 mm, 2.00 mm)
|r| = √(1.54² + 2²)
|r| = 2.52 mm
E = (9 x 10⁹ x 37.3 x 10⁻³)/(2.52 x 10⁻³)²
E = 5.287 X 10¹³ N/C
Thus, the electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.
Learn more about electric field here: brainly.com/question/14372859
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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
Answer:
θ = 10.28º
Explanation:
To find the angle of refraction use the equation of refraction
n₁ sin θ₁ = n₂ sin θ₂
where index 1 is for incident light and index 2 is for refracted light.
sin θ₂ = n₁ / n₂ sin θ
let's calculate
sin = 1 / 1.3 sin 0.23
sin = 0.175
θ= 0.17528 rad
let's reduce to degrees
θ = 0.17528 rad (180ª / pi rad)
θ = 10.28º
Answer:
The value is
Explanation:
From the question we are told that
The velocity which the rover is suppose to land with is
The mass of the rover and the parachute is
The drag coefficient is
The atmospheric density of Earth is
The acceleration due to gravity in Mars is
Generally the Mars atmosphere density is mathematically represented as
=>
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Generally the drag force on the rover and the parachute is mathematically represented as
=>
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Gnerally this drag force is mathematically represented as
Here A is the frontal area
So
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