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4 years ago

Describe how we can use scientific knowledge and reasoning to help guide us when making decisions

1 answer:

7 0

<span>A very simple example is the light bulb question. If the light bulb in your room goes out, scientific reasoning will first attribute it to a short circuit. Then, after setting up the hypothesis, we go and check if changing the fuse solves the problem. if not, the hypothesis is rejected and we change the hypothesis and do another experiment.</span>

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Consider an ideal intrinsic semiconductor in thermal equilibrium. No external forces or ﬁelds are applied to this semiconductor.

marissa [1.9K]

At temperatures above 0 K, the electron concentration in the conduction band is non-zero because some electrons from the valence band will overcome the bandgap by gained thermal energy. The correct option is D.

<h3>What are electrons?</h3>

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

Consider an ideal intrinsic semiconductor in thermal equilibrium. No external forces or ﬁelds are applied to this semiconductor. At temperatures above 0 K, the electron concentration in the conduction band is non-zero because some electrons from the valence band will overcome the bandgap by gained thermal energy.

The semiconductors are free from impurities. At room temperature. Some electrons ion valence band gains energy and cross the bandgap. They get transferred to the conduction band.

Thus, the correct option is D.

Learn more about electrons.

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5 0

2 years ago

You place 100 grams of ice, with a temperature of −10∘C, in a styrofoam cup. Then you add an unknown mass of water, with a tempe

Masteriza [31]

Answer:

Mass of water 2.9g

Explanation:

Ice

$m_{ice}=100g$

$c_{ice}=2J/g.K$

$T_{ice,initial}=-10\°C$

$T_{ice,final}=T_{equilibrium}=-5\°C$

Water

$c_{water}=4J/g.K$

$T_{water,initial}=10\°C$

$T_{water,final}=0\°C$

$T_{equilibrium}=-5\°C$

$l_{water}=300J/g$

$m_{water}=?g$

Step 1: Determine heat gained by ice

$Q_{ice}=m_{ice}c_{ice}(T_{ice,final}-T_{ice,initial})$

$Q_{ice}=100*2*(-5--10)$

$Q_{ice}=1000J$

Step 2; Determine heat lost by water

$Q_{water}=m_{water}c_{water}(T_{water,initial}-T_{water,final})+m_{water}l_{water}$

$Q_{water}=m_{water}*4*(10-0)+m_{water}*300$

$Q_{water}=40m_{water}+300m_{water}$

$Q_{water}=340m_{water}$

Step 3: Heat gained by ice is equivalent to heat lost by water

$Q_{ice}=Q_{water}$

$1000=340m_{water}$

$m_{water}=2.9g$

6 0

3 years ago

The ammeter below shows the current produced by a series of solar cells that contain zinc plates being used to power a simple se

nataly862011 [7]

The Ammeter shows: I = 15 A,
The circuit resistance is R = 0.2 Ohms.
By the Ohm`s Law:
R = V / I or:
V = I · R
V = 15 A · 0.2 Ohms
V = 3 V
Answer: C ) 3 V.

8 0

3 years ago

What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup

Fantom [35]

Answer:

3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²

- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²

Explanation:

Complete Question

3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?

- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.

The image of this setup attached to this question as obtained from online is attached to this solution.

Solution

3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.

Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²

The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²

- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.

I₂ = I₁ cos² θ

where

I₂ = intensity of light that passes through the second polarizer = ?

I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²

θ = angle between the major axis of the first and second polarizer = 30°

I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²

In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through

I₃ = I₂ cos² θ

I₃ = intensity of light that passes through the third/additional polarizer = ?

I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²

θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)

I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²

Hope this Helps!!!

5 0

3 years ago

You are ice skating on a rink in a friends backyard.you hit a stick that has frozen in the ice,and your skates stop suddenly.you

shusha [124]

That's because of inertia

4 0

3 years ago

Read 2 more answers