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3 months ago

What is alpha, beta and gamma radiation made of

Physics

1 answer:

algol133 months ago

6 0

Answer:

Alpha radiation is the name for the emission of an alpha particle in fact an helium nuclei, beta radiation is the emission of electrons or positrons , and gamma radiation is the term used for the emission of energetic photons.

Explanation:

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It is the final seconds of an ice hockey game between the Flyers and the Bruins. The Bruins are down by 1 point. With 20 s left

liberstina [14]

Answer:

$P =0.1094 \ m$ past the goal post

Yes the Flyers won the match

Explanation:

Generally the distance covered by the pluck during the last 1.25 seconds is mathematically represented as

$D = ut + \frac{1}{2} * a * t^2$

=> $D =30 * 1.25 + \frac{1}{2} * (-0.5) * (1.25)^2$

=> $D =37.1094 \ m$

Generally the position of the puck when the game clock reaches zero is mathematically represented as

$P = 37.1094 -37$

$P =0.1094 \ m$ past the goal post

Given that the Bruins where one point down and Flyers scored another goal it means that the flyer are now two point up hence they won the match

6 0

2 years ago

. Inside a conducting sphere of radius 1.2 m, there is a spherical cavity of radius 0.8 m. At the center of the cavity is a poin

MAVERICK [17]

Answer:

E = 1,873 10³ N / C

Explanation:

For this exercise we can use Gauss's law

Ф = E. dA = $q_{int}$ / ε₀

Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.

The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.

The surface of a sphere is

A = 4π r²

E 4π r² = q_{int} /ε₀

The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is

q_{int} = q₁ + q₂

q_{int} = (530 - 200) 10⁻⁹

q_{int} = 330 10⁻⁹ C

The electric field is

E = 1 / 4πε₀ q_{int} / r²

k = 1 / 4πε₀

E = k q_{int}/ r²

Let's calculate

E = 8.99 10⁹ 330 10⁻⁹/ 1.32²

E = 1,873 10³ N / C

7 0

2 years ago

Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri

irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

$d=\frac{2g(m1-m2)}{k}$

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$ (1)

With $K_{i}$ the initial kinetic energy, $U_{i}$ the initial potential energy, $K_{f}$ the final kinetic energy and $U_{f}$ the final potential energy. Note that initialy the masses are at rest so $K_{i} = 0$ , when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so $K_{f} =0$ . So, equation (1) becomes:

$U_{i}=U_{f}$ (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so $U_{i}=U_{gi}$ and at final moment we have potential gravitational energy and potential elastic energy so $U_{f}=U_{gf}+U_{ef}$ , using this on (2)

$U_{gi}=U_{gf}+U_{ef}$ (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), $U_{gi}=0$ and we have by (3) :

$0= U_{gf}+U_{ef}$ (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

$0=-m1gd+m2gd+\frac{kd^{2}}{2}$