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Ostrovityanka [42]
3 years ago
12

High energy waves have what

Physics
2 answers:
Hitman42 [59]3 years ago
5 0

Answer:

It means that the frequency will be a lot higher with a shorter wave.

GarryVolchara [31]3 years ago
3 0
High energy waves have Gamma rays
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The following are steps involved in transmission at the cholinergic synapse:
Westkost [7]

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

6 0
3 years ago
The ability to adjust to an artificially displaced or even inverted visual field is called
egoroff_w [7]

Answer:

méthode if séparation muddy water

3 0
2 years ago
A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar
Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

f=\mu mg

Put the value into the formula

f=0.8\times37

f=29.6\ N

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

f = \mu N

Put the value in to the formula

f=0.4\times37

f=14.8\ N

Hence, The frictional force acting on the block is 14.8 N.

6 0
3 years ago
At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6
Elis [28]

The particle has constant acceleration according to

\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k

Its velocity at time t is

\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du

\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t

\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k

Then the particle has position at time t according to

\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du

\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k

At at the point (3, 6, 9), i.e. when t=0, it has speed 8, so that

\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64

We know that at some time t=T, the particle is at the point (5, 2, 7), which tells us

\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}

and in particular we see that

v_{0y}=-2v_{0x}

and

v_{0z}=-v_{0x}

Then

{v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3

\implies v_{0y}=\mp\dfrac{8\sqrt6}3

\implies v_{0z}=\mp\dfrac{4\sqrt6}3

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k

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3 years ago
Describe the energy transformations that occur in a waterfall...
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3 years ago
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