All categories

2 years ago

What is the speed of alpha rays, beta rays and gamma rays ?

1 answer:

5 0

Alpha particles are released by proton rich unstable nuclei. It has no electrons to balance the protons making it positively charged particles moving at very high speeds. Gamma articles are electromagnetic waves of short wavelength and high frequency. It has a speed same as the speed of light. Beta particles are released by neutron rich unstable nuclei. It has a speed of less than the gamma particles.

You might be interested in

According to the Declaration, when do people have a right to revolution?

babunello [35]

Locke said that under natural law, all people have the right to life, liberty, and private property; under the social contract, the people could instigate a revolution against the government when it acted against the interests of citizens, to replace the government with one that served the interests of citizens.

3 0

2 years ago

In the periodic table, the most reactive metals are found a. in Group 1, the first column on the left. b. in Period 1, the first

Gnesinka [82]

Answer:

Bottom left corner for whatever group that is

Lithium, sodium, and potassium all react with water

3 0

3 years ago

Read 2 more answers

A small bolt with a mass of 41.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire

EleoNora [17]

Answer:

Amplitude will be equal to 0.091 m

Explanation:

Given mass of the slits = 41 gram = 0.041 kg

Frequency f = 1.65 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 1.65=10.362rad/sec$

Angular frequency is equal to $\omega =\sqrt{\frac{k}{m}}$

$10.362 =\sqrt{\frac{k}{0.041}}$

Squaring both side

$107.371 ={\frac{k}{0.041}}$

k = 4.40 N/m

For vertical osculation

$mg=kA$

$0.041\times 9.8=4.40\times A$

A = 0.091 m

So amplitude will be equal to 0.0391 m

8 0

3 years ago

The half-life of a certain isotope is 15 minutes. How much of a 400 g sample will remain after 90 minutes?12.5 g6.25 g26.7 g66.7

Virty [35]

There is a total of 6 half lives that need to take place.

ONE HALF LIFE = 200

TWO HALF LIFES = 100

THREE HALF LIFES = 50

FOUR HALF LIFES = 25

FIVE HALF LIFES = 12.5

SIX HALF LIFES = 6.25

The answer is 6.25g

4 0

3 years ago

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

3 years ago