Which of the following units are fundamental units in the S.I. system of measurement

salantis [7]

Answer:

Generally with successive half-life a new element is formed from the decayed nuclei:

isotope - daughter nucleus has same no. of protons (same atomic number) so the same element is formed in the decay

isotone - daughter nucleus has same number of neutrons so the atomic number has changed and a new element is formed (new atomic number)

isobar - daughter nucleus has same mass number - this could be an example of beta decay where the decayed nucleus has the same mass number but loses a neutron and gains a proton because of the lost electron and a new element is formed with the atomic number increases by 1

Example:

U238 Uranium - alpha to Th234

Th234 Thorium - 2 beta to U234 (back to U but atomic mass less by 4)

U234 Uranium - alpha to Th230

Th230 Thorium - alpha to Ra226

etc. but during beta decay (electron loss) a neutron is lost and a proton gained (answer a)

3 0

3 years ago

What does the thermosphere and stratmosphere have in common

Darya [45]

these include the troposphere (0 to 16 km), stratosphere (16 to 50 km), mesosphere (50 to 80km) and thermosphere (80 to 640km). The boundaries between these four layers are defined by abrupt changes in temperature, and include respectively the tropopause, stratopause and mesopause.

5 0

3 years ago

The motors that drive airplane propellers are, in some cases, tuned by using beats. The whirring motor produces a sound wave hav

sweet-ann [11.9K]

Answer:

A) 2 possible frequencies of second propellar = 424 rpm or 724 rpm

B) Correct frequency is f2 = 724 rpm

C) Reason is stated in explanation

Explanation:

A) We are given;

Frequency of first propeller; f1 = 574 rpm

Beat frequency; f_beat = 2.5 Hz = 2.5 × 60 rpm = 150 rpm

Now, formula for the beat frequency is;

f_beat = |f1 - f2|

Now, |f1 - f2| means it is inside an absolute value.

Thus, it means,

f1 - f2 = 150 or f2 - f1 = 150

Thus;

574 - f2 = 150

Or f2 - 574 = 150

So,f2 = 574 - 150 = 424 rpm or f2 = 574 + 150 = 724 rpm

2 possible frequencies of second propellar = 424 rpm or 724 rpm

B) Now, if we increase the speed of the second propellar slightly, it means that f2 will increase as well.

Now, from the 2 possible values of f2 gotten, we can see that for f1 - f2 = 150, when we increase f2, the beat frequency will reduce while for f2 - f1 = 150, when we increase f2, the beat frequency will increase.

Thus, it's the frequency of f2 gotten in f2 - f1 = 150 that is the correct answer.

Thus, when we increase the speed of the second propellar slightly, it means that;

f2 = 724 rpm

C) Answer in part B is correct because as we increase the speed of the second propellar, the frequency will also increase and the value of f2 that corresponds with an increase in speed is 724 rpm.

7 0

3 years ago

Fill in the blank. When the northern hemisphere experiences autumn, the southern hemisphere experiences __________.

Helga [31]

Answer:

When the northern hemisphere experiences autumn, the southern hemisphere experiences spring

Explanation:

When the northern hemisphere experiences autumn, the southern hemisphere experiences spring

This is due to the equinoxes. An equinox is an event that occurs twice a year. During these seasons all areas of the Earth's surface experience an equal amount of daylight and darkness. The sun is on the equator line, so the day and night in both hemispheres have the same duration. At that time the part of the Earth closest to the Sun is the equator.

The shape of the Earth means that while the spring equinox is experienced in the northern hemisphere, the autumn equinox is entered in the southern hemisphere.

8 0

4 years ago

Read 2 more answers

An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

irinina [24]

Answer:

a) $\sigma_{\rm in} = -2.18~{\rm \mu C/m^2}$

b) $\sigma_{\rm out}= 1.12~{\rm \mu C/m^2}$

c) $E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

$Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}$

$Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}$

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

$Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}$

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

$\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}$

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

$\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}$

c) The electric field outside the cylinder can be found by Gauss' Law:

$\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

$E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

4 0

3 years ago