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3 years ago

Explain how the information from the stimuli is communicated to the brain.

Physics

2 answers:

irakobra [83]3 years ago

6 0

Answer:

Answer:Neurons communicate through an electrochemical process. Sensory receptors interact with stimuli such as light, sound, temperature, and pain which is transformed into a code that is carried to the brain by a chain of neurons. Then systems of neurons in the brain interpret this information.

bro edit it yrself

dedylja [7]3 years ago

4 0

Answer:

Information is carried when the sensory receptors interact with the stimuli. Gathered information like sound, temperature, pain, etc. gets transferred into code that gets transferred to the brain.

Explanation:

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Three capacitors having capacitances of 8.40, 8.40, and 4.20 μFμF, respectively, are connected in series across a 36.0-V potenti

son4ous [18]

Answer:

a)Q=71.4 μ C

b)ΔV' = 10.2 V

Explanation:

Given that

C ₁= 8.7 μF

C₂ = 8.2 μF

C₃ = 4.1 μF

The potential difference of the battery, ΔV= 34 V

When connected in series

1/C = 1/C ₁ + 1/C₂ + 1/C₃

1/ C= 1/8.4 +1 / 8.4 + 1/4.2

C=2.1 μF

As we know that when capacitor are connected in series then they have same charge,Q

Q= C ΔV

Q= 2.1 x 34 μ C

Q=71.4 μ C

As we know that when capacitor are connected in parallel then they have same voltage difference.

Q'= C' ΔV'

C'= C ₁+C₂+C₃ (For parallel connection)

C'= 8.4 + 8.4 + 4.2 μF

C'=21 μF

Q'= C' ΔV'

Q'=3 Q

3 x 71.4= 21 ΔV'

ΔV' = 10.2 V

3 0

2 years ago

Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of

kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

$m=\dfrac{F}{g}$

$m=\dfrac{160\ lb}{32\ ft/s^2}$

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

$m=\dfrac{F}{g}$

$m=\dfrac{1.9\ lb}{32\ ft/s^2}$

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

$W=2300\ kg\times 32\ ft/s^2=73600\ lb$

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

$W=0.022\ kg\times 32\ ft/s^2=0.704\ lb$

Hence, this is the required solution.

5 0

2 years ago

A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w

FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$

As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

$a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2$

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

$E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C$

⇒ E = 1.04*10⁻¹ N/C

5 0

3 years ago

Why might you want to use a single fixed pulley?

hjlf

A single fixed pulley can be used to raise or lower lightweight objects.

Option b

Explanation:

A pulley is a simple machine tool which is used to make lifting or lowering tasks easy. A single fixed pulley is a system involving only one pulley fixed on a constant rigid support with a rope wrapped around the wheel. Such a system can be used only to change the direction of applied force in raising or lowering small, lightweight objects which need minimal work force.

A single fixed pulley system helps only in redirecting the applied force direction by using a rope and wheel assembly. The work done in such a case remains the same and hence it is not preferred to use it in lifting heavy objects. Neither is the required force reduced in case of a single fixed pulley system. A movable pulley helps in achieving (A) and (C).

4 0

2 years ago

A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released

statuscvo [17]

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

mass of the block, m = 2.4 kg
height of the block, h = 5 m
compression of the spring, x = 25 cm = 0.25 m

The spring constant is calculated as follows;

$F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m$

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

$\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2 - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2 = mgh - \frac{1}{2} kx^2\\\\mv^2 = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s$

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

8 0

2 years ago