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PSYCHO15rus [73]
2 years ago
5

Atomic bombs send out a shock wave when they are detonated. This occurs because the bomb's detonation superheats the air particl

es nearby, which cause them to bump into other air particles that further the wave.
Physics
2 answers:
vova2212 [387]2 years ago
7 0

Question:

Atomic bombs send out a shock wave when they are detonated. this occurs because the bomb’s detonation superheats the air particles nearby, which cause them to bump into other air particles that further the wave.

which statement can most likely be made about the shock waves of atomic bombs?

a. they are a mechanical waves.

b. they are an electromagnetic waves.

c. they will act like x-rays.

d. they will act like light rays

Answer:

The correct option is;

a. they are a mechanical waves.

Explanation:

The shock waves of atomic bombs are mechanical waves which constitute the movement, back and forth, of particulate mater (oscillation), which results in energy (heat) transfer by means of a medium (air particles) over a distance.

As it is a longitudinal, mechanical wave, the motion of the air particles is in the direction of the wave and is limited to within the original region of the particle before the blast, maintaining as equilibrium position while transferring the energy to the adjacent neighboring air particle.

Dahasolnce [82]2 years ago
6 0

Answer:

That's right: The initial energy emission occurs by 80% or more in the form of gamma rays but these are quickly absorbed and dispersed mostly by air in little more than a microsecond, converting gamma radiation into thermal radiation (thermal pulse ) and kinetic energy (shock wave) which are actually the two dominant effects in the initial moments of the explosion. The rest of the energy is released in the form of delayed radiation (fallout or fallout) and is not always counted when measuring the performance of the explosion.

Explanation:

High altitude explosions produce greater damage and extreme radiation flux due to lower air density (photons encounter less opposition) and consequently a higher blast wave is generated.

For a long time before the invention of the bomb, some scientists believed that its detonation on the surface could cause the Earth's atmosphere to ignite, generating a global chain reaction in which nitrogen atoms would join together to form carbon and oxygen. This fact soon proved impossible since the densities necessary for these reactions to take place have to be much higher than atmospheric ones, and although there may be additional fusion reactions at the heart of the explosion, they do not provide enough energy to amplify and spreading the nuclear reaction to the rest of the atmosphere and the production of heavy elements ceases immediately. Regardless, this idea persists today as a misunderstanding rumor among many people.

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Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,
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Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F = \frac{k*q_1*q_2}{d^2} Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)  

d: distance between the charges in meters (m)

Equivalence  

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

F_{13} = \frac{k*q_1*q_3}{d_1^2}

F_{23} = \frac{k*q_2*q_3}{d_2^2}

F₁₃ = F₂₃

\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2} We eliminate k and q₃ on both sides

\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}

\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}

\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2} We eliminate 10⁻⁶ on both sides

(1-x)^2 = \frac{7}{5} x^2

1-2x+x^2=\frac{7}{5} x^2

5-10x+5x^2=7 x^2

2x^2+10x-5=0

We solve the quadratic equation:

x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m

x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in  opposite way .

x = 0.458m = 45.8cm

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Answer:

Explanation:

a) 1.00 - 0.12 = 0.88

m = 1200(0.88)^t

b) t = ln(m/1200) / ln(0.88)

c) m = 1200(0.88)^10 = 334.20 g

d) t = ln(10/1200) / ln(0.88) = 37.451... = 37 s

e) t = ln(1/1200) / ln(0.88) = 55.463... = 55 s

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A microwave oven operates at 2.00 ghz . what is the wavelength of the radiation produced by this appliance? express the waveleng
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Answer:

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Explanation:

The object distance is y= 6in.

Because the surface is flat, the radius of curvature is infinity .

The incident index is n_i=\frac{4}{3} and the transmitted index is n_t= 1.

The single interface equation is \frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}

Substituting the quantities given in the problem,

\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0

The image distance is then y^i=-\frac{18}{4}in =-4.5in

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