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boyakko [2]
4 years ago
6

What is a peppered moth?

Physics
2 answers:
amm18124 years ago
6 0

The peppered moth is a temperate species of night-flying moth. Peppered moth evolution is an example of population genetics and natural selection.
Natali [406]4 years ago
4 0
<span>The Peppered Moth is widespread in Britain and Ireland and frequently found in ordinary back gardens, yet its amazing story has made it famous all over the world. It is one of the best known examples of evolution by natural selection, Darwin's great discovery, and is often referred to as 'Darwin's moth'. 

hope it helps!</span>
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The explanation of the relationship between load and extension is not in Hooke's Law.
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Accorrding to Kepler's third law, a planet whose distance from the Sun is 2 A.U. would have an orbital period of how many Earth-
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Explanation:

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3 years ago
A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in
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The answer:
the full question is as follow:
 <span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D? 

As shown in the figure, 
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
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let  W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics

Magnitude
 A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
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3 years ago
At the circus, a 100.-kilogram clown is fired at 15 meters per second from a 500.- kilogram cannon. What is the recoil speed of
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The recoil velocity of cannon is (4) 5.0 m/s

Explanation:

We can find the recoil velocity from the law of conservation of momentum.

The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.

Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec

m1 be clown's mass = 100kg and m2 be cannon's mass given = 500kg.

So recoil velocity of cannon v2 is given by,

v2 = -(m1÷m2)v1

v2 = -(100÷500)15

v2 = -5 m/s

where the minus sign refers to the direction of cannon's recoil velocity being opposite to that of clown.

Hence, option (4)5.0 m/s is the correct answer.

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