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2 years ago

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An explosion at ground level leaves a crater with a diameter that is proportional to the energy of the explosion raised to the 1

/3 power; an explosion of 1 megaton of TNT leaves a crater with a 1 km diameter. An ancient impact crater is found with a 36 km diameter. What was the kinetic energy associated with that impact, in terms of (a) megatons of TNT (1 megaton yields 4.2 × 1015 J) and (b) Hiroshima bomb equivalents (13 kilotons of TNT each)?

2 answers:

adoni [48]2 years ago

8 0

Answer:

a) 3.024*10^12 MT

b) 2.326*10^14 units

Explanation:

Given:

- The relationship between diameter d of the crater and energy E is:

d = k*E^(1/3)

- Where k is constant of proportionality

- Explosion of 1 MT of TNT creates 1 km diameter.

Find:

An ancient impact crater is found with a 36 km diameter

(a) megatons of TNT (1 megaton yields 4.2 × 1015 J)

(b) Hiroshima bomb equivalents (13 kilotons of TNT each)?

Solution:

- To evaluate constant k:

1,000 = k*(4.2*10^15)^(1/3)

k = 1.543*10^-5

- The relationship can now be expressed as:

d = 1.543*10^-5*E^(1/3)

- For a crater of d = 36 km, the associated energy is:

E = (36,000 / 1.543*10^-5)^3

E = 1.2699*10^28 J

a)

1 MT of TNT ---------------------> 4.2 × 10^15 J

x ---------------------> 1.2699*10^28 J

x = 1.2699*10^28 /4.2 × 10^15 = 3.024*10^12 MT

b) The E = 3.024*10^12 MT.

1 unit of Hiroshima bomb is = 0.013 MT of TNT

Units of bomb = 3.024*10^12 MT / 0.013 MT = 2.326*10^14 units

vlada-n [284]2 years ago

6 0

Answer:

a)125 GT

b)9615384.6 Hiroshima bombs

Explanation:

We know that

D=k(E)^(1/3) (and I will keep D in km and E in MT of TNT)

D= diameter

Where k is a constant of proportionality.

We also have

1=k*(1)^(1/3), k=1

(a) For D=50, E=(D^3), so E= 125000 MT or 125 GT

(b) For Hiroshima equivalents, that's 125000 MT / 13 KT = 125000 MT / 0.013 MT = 9615384.6 Hiroshima bombs.

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Answer:

Option B is the correct answer.

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h= 161.06 m

Explanation:

Given that

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Two speakers, A and B, produce identical sound waves. A listener is 3.2 m away from speaker A. The listener finds the lowest fre

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Answer:

0.83 m or 5.57 m

Explanation:

Destructive interference will occur when the distances from the speakers differ by 1/2 wavelength.

The length of 1 cycle of 72.4 Hz is ...

λ = v/f = (343 m/s)/(72.4 Hz) ≈ 4.738 m

So, the distance of the listener from speaker B is ...

3.2 m ± (4.738 m)/2 = {0.83 m, 5.57 m} . . . either of these distances

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The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

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<em>Comment on the speed of sound</em>

We don't know what speed you are to use for the speed of sound. We have used 343 m/s. Some sources use 340 m/s, which will give a result different by 2 or 3 cm.

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To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

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According to the data given we have to,

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