Ask question

Ask question

All categories

4 years ago

3

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]4 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]4 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.

You might be interested in

What is the formula for calculating the efficiency of a heat engine?

Olegator [25]

Explanation:

The efficiency of a heat engine is defined as the ratio of the work output to the heat input into the system.

It is highly impossible to get an engine that is 100% efficient.
A heat engine that is 100% efficient suggests that all the heat input into the system is used to do work completely.
There is no heat loss.

Efficiency = $\frac{work output}{heat input}$ x 100

Efficiency has no unit

Learn more:

Heat brainly.com/question/914750

#learnwithBrainly

7 0

3 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c

EleoNora [17]

Answer:

63.8 V

Explanation:

We are given that

$A_1=1.6 cm^2=1.6\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

$A_2=1.2 cm^2=1.2\times 10^{-4} m^2$

$A_3=4.4 cm^2=4.4\times 10^{-4} m^2$

$A_4=7 cm^2=7\times 10^{-4} m^2$

Potential difference,V=140 V

We know that

$R=\frac{\rho l}{A}$

According to question

$l_1=l_2=l_3=l_4=l$

In series

$R=R_1+R_2+R_3+R_4$

$R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})$

$R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})$

$R=\rho l(18284.6)$

$I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}$

Potential across 1.2 square cm= $V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V$

Hence, the voltage across the 1.2 square cm wire=63.8 V

3 0

3 years ago

How does the angle of incident and reflection affect your solar oven?

JulijaS [17]

It determines whether the light beams will be concentrated or scattered.

7 0

3 years ago

Conditional probability involves events that are independent. true or false?

kiruha [24]

Its false I got It right .

3 0

3 years ago

Which of the following is not a transverse wave?A) soundB) lightC) radioD) all of theseE) none of these

quester [9]

The answer is light, because a transverse wave is a wave vibrating at right angles to the direction of its propagation

7 0

3 years ago