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polet [3.4K]
2 years ago
12

Which of the following graphs shows the relationship between two variables that obey the inverse square law?​

Physics
1 answer:
blondinia [14]2 years ago
8 0

Well, if we've been paying attention in class, we already KNOW that the electrostatic force changes as the inverse square of the distance, and the top graph is conveniently labeled "Electrostatic Force".

But if we didn't already know that, we'd have to examine the graphs, and find the one where 'y' changes like 1/x² .

The top graph does that.  After 1 unit of time, the force is 350.  Double the time to 2 units, and the force should drop to 1/4 of 350 ... sure enough, it's a little less than 90.  Double the time again, to 4 units, and it should drop to 1/4 of a little less than 90 ... by golly, it's down below 30.

The first graph is what an inverse square looks like.  Now that you've worked out this graph, you'll know an inverse square relationship whenever you see it.  

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What is a pesticide?
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A pesticide is something that is used to kill / deter pests that eat / destroy crop.

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1 year ago
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A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the studen
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We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I  attached we can see that:
\frac{h_i}{h_0}=\frac{d_i}{d_0}
Two triangles that I marked are similar and from this we get:
\frac{h_i}{h_0}=\frac{d_i-f}{f}
The image and the object must have the same height so we get:
\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\
1=\frac{d_i-f}{f}\\
d_i=2f
This tells how far the screen should be from the lens. 
The position of the screen on the optical bench is:
S=20.5cm+2f=20.5+2\cdot 15cm=50.5cm

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1 year ago
The model of the atom has changed as scientists have gathered new evidence. Four models of the atom are shown below, but one imp
nexus9112 [7]

Answer: Dalton’s model

Explanation:

In the attached image we can see four atomic models labeled with four letters:

W represents the current and accepeted atomic model: a nucleus with an electron cloud, where the orbit and position of the electrons around the nucleus is defined by specific regions (associated with specific energy levels) where there is a greater probability of finding the electron at any given moment. It is important to note this model was improved by the works in quantum physics done by Louis de Broglie and Erwin Schrodinger.

X represents Rutherford's model (This model was proposed after Thomson's model). Ernest Rutherford conducted a series of experiments in order to corroborate Thomson's atomic model. However the results of the experiment led him to find out there is a concentration of charge in the atom's core (which was later called nucleus) surrounded by electrons.  This lead to a new atomic model, in which the atom has a positive charged nucleus surrounded by negative charged particles that move similar to the orbit of the planet around the Sun.

Y represents Thomson's model, also called  the <em>plum pudding</em> model. This scientific found out that atoms contain small subatomic particles with a negative charge (later called electrons). However, taking into consideration that at that time there was still no evidence of the atom nucleus, Thomson thought the electrons were immersed in the atom of positive charge that counteracted the negative charge of the electrons. Just like the raisins embedded in a pudding or bread.

Z represents Bohr's model. This model was proposed by the danish physicist Niels Bohr after Rutherford's model. In fact, this model was Rutherford's model with the following addition: electrons orbit the nucleus (like planets around the sun) in specific orbits at different energy levels around the nucleus.

So, the only missing model is <u>Dalton's model</u>, which was the first atomic model:  the atom represented as a solid, indestructible and indivisible mass. An idea that was already accepted by that time since the ancient Greeks.

4 0
2 years ago
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Imagine that a small boy and his much larger father are enjoying a winter morning, sliding along the ice on a nearby playground.
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You can write an hypothesis such as this:
The weight of an object has effects on the operating frictional force, the greater the weight, the higher the operating frictional force.
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An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst
sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

T_1 = 500°C

T_2 = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to T_{avg} = 535.5K \approx 550K

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

P_r = 0.63

A)

Number for the first strips is equal to

R_e_x = \frac{u_o.L}{v}

R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4

Calculating heat transfer coefficient from the first strip

h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3

h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2

The rate of convection heat transfer from the first strip is

q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W

The rate of convection heat transfer from the fifth trip is equal to

q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)

h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2

Calculating h_o_-_4

h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2

The rate of convection heat transfer from the tenth strip is

q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)

h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2

Calculating

h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2

Calculating the rate of convection heat transfer from the tenth strip

q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W

The rate of convection heat transfer from 25th strip is equal to

q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)

Calculating h_o_-_2_5

h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2

Calculating h_o_-_2_4

h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2

Calculating the rate of convection heat transfer from the tenth strip

q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W

6 0
2 years ago
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