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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 :

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

You might be interested in

GIVING BRAINLIEST Which statement best explains how Earth's magnetic field is generated? Earth's core spins slower than its crus

zimovet [89]

Answer:

The convection of heat within the mantle Rotation of liquid iron in the outer core

Explanation:

The electrical currents and convecting movements within the Earth's liquid core generate the planet's powerful magnetic field, which in turn drives the inner core in the way that can now be observed.

8 0

3 years ago

find the average velocity of a bicycle that starts 100km south and is 120km south of town after 0.4 hours

dusya [7]

Answer:

The average velocity is 50 km/h south

Explanation:

The average velocity of an object is its total displacement divided by

the total time taken.

That means it is the rate at which an object changes its position from

one place to another.

Average velocity is a vector quantity.

The SI unit is meters per second.

A bicycle that starts 100 km south and is 120 km south of town after

0.4 hour.

The displacement = 120 - 100 = 20 km south

The time = 0.4 hour

The average velocity = $\frac{D}{T}$ , where D is the displacement

and t is the time

The average velocity of the bicycle = $\frac{20}{0.4}=50$ km/h

<em>The average velocity is 50 km/h south</em>

If you want it in meter per second, change the kilometer to meter

and change the hour to seconds

1 km = 1000 m

1 hour = 60 × 60 = 3600 seconds

The average velocity of the bicycle = $\frac{50(1000)}{3600}=13.89$ m/s south

5 0

3 years ago

An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$

Now we can calculate the acceleration of the plane, by using Newton's second law:

$a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2$

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

$v^2- u^2 = 2aS$

where

$v=?$ is the final velocity

$u=66.0 m/s$ is the initial velocity

$a=1.69 m/s^2$ is the acceleration

$S=5.00 \cdot 10^2 m$ is the distance travelled

Solving for v, we find

$v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s$

8 0

3 years ago

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b

Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $\frac{1}{2} kx^{2} +PE1$