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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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A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio
xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string  

For without breaking the string tension must be equal to the centripetal force

So T=\frac{mv^2}{r}

So 350=\frac{0.5\times v^2}{1}

v^2=700

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

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Our Solar System's planets formed when fragments in space joined together to form growing spheres in a process known as ____​
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4 0
2 years ago
Read 2 more answers
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

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2 years ago
Atoms of two different elements must have different
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Explanation:

C. Atomic numbers....

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