What is the delay between a stimulus and the response it triggers in an organism?

What is an non example of Cultural diffusion

belka [17]

Beliefs is a non example

4 0

2 years ago

The volume of a gas can be converted to moles by

BaLLatris [955]

B. Because I say so

6 0

2 years ago

How can I convert this?<br>Please answer with solution. Thank you.

fomenos

Answer:

1 hr 45 min

Explanation:

8 0

2 years ago

You drop your frozen rock from a green bridge. The frozen rock starts from rest (initial velocity = 0ms). The rock takes 4.3s to

valentinak56 [21]

Answer:

The velocity of the frozen rock at $t = 1.5\,s$ is -14.711 meters per second.

Explanation:

The frozen rock experiments a free fall, which is a type of uniform accelerated motion due to gravity and air viscosity and earth's rotation effect are neglected. In this case, we need to find the final velocity ( $v$ ), measured in meters per second, of the frozen rock at given instant and whose kinematic formula is:

$v = v_{o} + g\cdot t$ (Eq. 1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$g$ - Gravity acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we get that $v_{o} = 0\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $1.5\,s$ , then final velocity is:

$v = 0\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (1.5\,s)$

$v = -14.711\,\frac{m}{s}$

The velocity of the frozen rock at $t = 1.5\,s$ is -14.711 meters per second.

5 0

2 years ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

3 0

2 years ago