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AfilCa [17]
3 years ago
14

17.

Physics
1 answer:
avanturin [10]3 years ago
6 0

Answer:

Gamma rays

Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies.

Explanation:

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When Missourians carve their initials into the bark of a tree, the damage leaves the tree open to ____________________.
sasho [114]

Answer:

In fact, carving letters into a tree probably won't hurt it. ... In general, the tree will compartmentalize the wound and it will heal over. The initials that remain visible are essentially scar tissue, permanent scar tissue.

Explanation:

Unfortunately, when carving into the trunk of a tree the blade of a knife often penetrates the outer bark and cuts into the inner bark. ... In cases that the phloem is damaged all the way around the trunk (in a ring for example), the tree will slowly and eventually starve to death.

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luke_raines19

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3 years ago
Pour chacune des réactions chimiques
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2 years ago
9.58 A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. T
andrey2020 [161]

Answer:

The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.

See attachment below for full solution

Explanation:

This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.

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3 years ago
Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person
sertanlavr [38]

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-v_{s})

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

3 0
3 years ago
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