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3 years ago

In beta decay, what is emitted?

2 answers:

8 0

An electron is emitted in both positive and negative beta decay, although the positive one is called positron emission.

bagirrra123 [75]3 years ago

4 0

Answer:The correct answer is an electron.

Explanation:

Beta decay : In this process, a neutron is converted into a proton and an electron. In this decay, the atomic number is increased by 1 unit.

General representation of beta decay :

$_Z^A\textrm{X}\rightarrow _{Z+1}^{A}Y+_{-1}^0e$

Where as:

Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.

General representation of alpha decay :

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}Y+_2^4\alpha$

Gamma decay : In this process, the unstable nuclei transformed into stable nuclei by releasing gamma rays. In this process, the mass number remains same.

General representation of gamma decay :

$_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma$ .

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Which is the currrent SI unit to use in measuring the amount of material contained in a solid sphere

olya-2409 [2.1K]

Hey

SI unit for measure the amount of material contained in a solid sphere metre cube(m^3).

mean it have Volume .

3 0

4 years ago

Compare and contrast electric conductors and electric insulators. Give an example of each.

ludmilkaskok [199]

An example of conductors of heat would be iron pans. a example of electric insulators would be copper, gold and silver. to contrast conductors and insulators, insulators let electricity pass through them while conductors restricts electricity. both conductors and insulators can work with lithium and sodium.

3 0

3 years ago

QUESTION 1 The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 968 Hz?

Anton [14]

Answer:

Explanation:

340 m/s / 968 cyc/s = 0.3512396... ≈ 35.1 cm

7 0

3 years ago

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$

$t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s$

So, the total time is:

$t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s$

7 0

4 years ago

A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su

aniked [119]

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒ G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² = 4 ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

7 0

3 years ago