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Firdavs [7]
3 years ago
9

¿Qué pasa en nuestro cerebro cuando tomamos decisiones? Ejemplo: Estoy tocando una plancha caliente, y saco la mano rápidamente,

¿que paso en mi cerebro?
Physics
1 answer:
Natalija [7]3 years ago
4 0

Answer:

Debido a la información de las neuronas sensoriales al cerebro.

Explicación:

Cuando toco una plancha caliente, extiendo la mano rápidamente porque las neuronas sensoriales presentes en la piel de nuestro cuerpo envían señales al sistema nervioso central, es decir, al cerebro. El cerebro toma decisiones de acuerdo con los estímulos y envía instrucciones a través de las neuronas motoras a los músculos para retirar las manos del hierro caliente a fin de evitar efectos nocivos en la piel debido al calor. El lóbulo frontal es una parte del cerebro que controla la personalidad, la toma de decisiones y el razonamiento.

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Svetllana [295]

Answer:

Beta 17,000K, bc warmer is more blue

5 0
3 years ago
What is the SPEED of a ball that travels 120 m in 6s?
AVprozaik [17]

Answer:

20m/second

Explanation:

The reason the answer is 20m/second is because to find the speed of the ball  in this question you have to divide the distance over the time giving you the result of 20m/second

7 0
3 years ago
4. Lead has a density of 11.5g/cmº. A rectangular block of lead measures 7cm x5cmx2cm.
lisov135 [29]

Answer:

(a) 70cm³

(b) 805 grams

Explanation:

(a) V = L×B×H

= 7cm×5cm×2cm

= 35cm×2cm

= 70cm³

(b) Mass = Volume × Density

= 70cm³ × 11.5g/cm³

= 805 grams

3 0
3 years ago
A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut
blsea [12.9K]

Answer:

1.E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2.E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

\rho (r) = \alpha (1-\frac{r}{R})

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

\int\, da = 2\pi r h

where ‘h’ is the length of the imaginary Gaussian surface.

Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3. At the boundary where r = R:

E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}

As can be seen from above, two E-field values are equal as predicted.

4 0
3 years ago
Mimas, a moon of Saturn, has an orbital radius of 1.62 × 108 m and an orbital period of about 23.21 h. Use Newton’s version of K
Drupady [299]

Answer:

3.60432\times 10^{26}\ kg

Explanation:

a = Orbital radius = 1.62\times 10^8\ m

T = Orbital period = 23.21 hours

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

From Kepler's third law we get

M=\frac{4\pi^2a^3}{GT^2}\\\Rightarrow M=\frac{4\pi^2\times (1.62\times 10^8)^3}{6.67\times 10^{-11}\times (23.21\times 3600)^2}\\\Rightarrow M=3.60432\times 10^{26}\ kg

From the given data the mass of Saturn is 3.60432\times 10^{26}\ kg

8 0
3 years ago
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