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Rom4ik [11]
3 years ago
8

This design enabled them to control for several factors. The total area of the patch plus the corridor, or the patch plus the wi

ngs was the same. The total amount of edge (perimeter of each patch) was similar between the two treatments. Both patch types were the same distance from the central patch.
Physics
1 answer:
anyanavicka [17]3 years ago
7 0

Answer:

Graph 1: This design does not control for distance from the central patch.

Graph 2: This design does not control for area of the patch.

Graph 3: In this design, there is no difference between the treatments.

Graph 4: This design does not control for edge effects.

Explanation:

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The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a
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The intensity on a screen 20 ft from the light will be 0.125-foot candles.

<h3>What is the distance?</h3>

Distance is a numerical representation of the length between two objects or locations.

The intensity I of light varies inversely as the square of the distance D from the source;

I∝(1/D²)

The ratio of the intensity of the two cases;

\rm \frac{I_1}{I_2} =(\frac{D_2}{D_1} )^2\\\\ \rm \frac{2}{I_2} =(\frac{20}{5} )^2\\\\ \frac{2}{I_2} =4^2 \\\\ I_2= \frac{2}{16} \\\\  I_2= 0.125 \ foot-candles

Hence, the intensity on a screen 20 ft from the light will be 0.125 foot-candles

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2 years ago
Why is pure oxygen stored as a liquid under pressure
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<h2>Answer: It is highly flammable.</h2>

Explanation:

Liquid oxygen is created from oxygen atoms that have been forced to assume the liquid state due to <u>compression (change of pressure) and temperature modification. </u>

Specifically this is achieved by cooling the oxygen enough to change it to its liquid state. So,<u> as the temperature drops, the atoms move more slowly because they have less energy. </u>

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Why the center of Newton rings is dark?​
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The point of contact the path difference is zero but one of the interfering ray is reflected so the effective path difference becomes λ/2 thus the condition of minimum intensity is created in the center.
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4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc
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a) 15.2 m/s^2

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

- The forward force of 3.55 N

- The backward, resistive force of 0.52 N

So, the net force forward is

\sum F=3.55-0.52=3.03 N

Now we can find the acceleration by using Newton's second law of motion, which states that:

\sum F=ma

where

m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2

b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

W=mg

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

g=9.8 m/s^2 is the gravitational field strength

So, the gravitational force on the box is

W=(0.200)(9.8)=1.96 N

c)

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

\sum F=N-W

According to Newton's second law,

\sum F=ma

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

a=0

This means that the net force is zero:

\sum F=0

And so, we can find the normal force:

N-W=0\\N=W=1.96 N

4 0
3 years ago
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