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3 years ago

12

Why is alpha rays not suitable to scan baggage at the airport?

1 answer:

Inga [223]3 years ago

4 0

Answer:

Alpha particles are not used to check baggage in airport as these rays are not penetrated far through matter.

<u>Explanation: </u>

Alpha particles are produced by the <em>decay of a radioactive nucleus.</em> These alpha rays are <em>ceased by paper</em>. The <em>nucleus</em> of the helium atom is made up of <em>two protons and two neutrons</em> is the particle <em>ejected by alpha particle</em>.

As helium is <em>harmless and as an inert gas</em>, so these particles are not dangerous in themselves. The airport baggage security uses only X-ray scanners to inspect. These rays identify the <em>hidden items inside the baggage. </em>

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A golf club consists of a shaft connected to a club head. The golf club can be modeled as a uniform rod of length ℓ and mass m1

slega [8]

Answer:

m₁ (R +ℓ/2) / /( m₁ + m₂ )

Explanation:

We shall consider centre of the club's head as origin .

The centre of mass of the rod attached with sphere will be away from the centre of sphere by a distance

= R + ℓ /2 and its mass is m₁

mass of sphere is m₂ and distance of its centre of mass from origin is zero

So required CM = m₁ x₁ + m₂ x₂ / ( m₁ + m₂ )

= m₁ (R + ℓ /2) + m₂ x 0 /( m₁ + m₂ )

= m₁ (R +ℓ/2) / /( m₁ + m₂ )

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3 years ago

In a particle accelerator, a magnetic force keeps a lithium nucleus (mass 6.02 u) traveling in a circular path with a radius of

klasskru [66]

Answer:

The value is $F_B = 1.074 *10^{-11} \ N$

Explanation:

From the question we are told that

The mass of the lithium nucleus is $m = 6.02 \ u = 6.02 * 1.66*10^{-27} = 9.9932*10^{-27}\ kg$

The radius is $r = 0.536 \ m$

The speed of the lithium nucleus is $v = 0.08 c = 0.08 * 3.0*10^{8} = 2.40*10^{7} \ m/s$

Given that the lithium nucleus is travelling in a circular path, the magnetic force will be equivalent to a centripetal force so

$F_B = F_C$

And $F_C[tex] is centripetal force mathematically represented as [tex]F_C = \frac{m * v^2}{r}$

So

$F_B = F_C = \frac{m * v^2}{r}$

=> $F_B = F_C = \frac{9.9932*10^{-27} * [2.40*10^{7}]^2}{ 0.536}$

=> $F_B = 1.074 *10^{-11} \ N$

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4 years ago

Eugene is studying the levels of structural organization of an animal’s body. Which level would describe a dog’s eye

mrs_skeptik [129]

The levels of structural organization:
cell ---> tissue ---> organ ---> organ system ---> organism

A dog's eye would be an organ.

6 0

4 years ago

Read 2 more answers

How much force is needed to push a 20kg at 0.2 m/s/s<br> A)10 N<br> B)20 N<br> C)4 N<br> D)40 N

ad-work [718]

Answer:

4

Explanation:

20x.2=4

4 0

3 years ago

A 3-kg ball is thrown with a speed of 8 m/s at an unknown angle above the horizontal. The ball attains a maximum height of 2.8 m

Alina [70]

Answer:

13.5 J

Explanation:

mass of ball, m = 3 kg

maximum height, h = 2.8 m

initial speed, u = 8 m/ s

Angle of projection, θ

use the formula of maximum height

$H = \frac{u^{2}Sin^{2}\theta }{2g}$

$2.8 = \frac{8^{2}Sin^{2}\theta }{2\times 9.8}$

Sin θ = 0.926

θ = 67.8°

The velocity at maximum height is u Cosθ = 8 Cos 67.8 = 3 m/s

So, kinetic energy at maximum height

$K=\frac{1}{2}mv^{2}$

K = 0.5 x 3 x 3 x 3

K = 13.5 J

8 0

4 years ago