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2 years ago

What elements are not a major category of elements on the periodic table

Physics

1 answer:

lana66690 [7]2 years ago

3 0

Answer:

Group and Periods of the Periodic Table of Elements. The three major groups on the Periodic Table are the metals, nonmetals and metalloids. Elements within each group have similar physical and chemical properties.

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The weld bead in SMAW is formed by the?

SVETLANKA909090 [29]

The answer is C I just took this quiz.

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2 years ago

On average, how many stars would we have to search before we would expect to hear a signal? assume there are 500 billion stars i

Keith_Richards [23]

We would have to search at least 5,000,000,000 (5 billion) stars before we would expect to hear a signal.

To find out the number of stars that we will need to search to find a signal, we need to use the following formula:

total of stars/civilizations
500,000,000,000 (500 billion) stars / 100 civilization = 5,000,000,000 (5 billion)

This shows it is expected to find a civilization every 5 billion stars, and therefore it is necessary to search at least 5 billion stars before hearing a signal from any civilization.

Note: This question is incomplete; here is the complete question.

On average, how many stars would we have to search before we would expect to hear a signal? Assume there are 500 billion stars in the galaxy.

Assuming 100 civilizations existed.

Learn more about stars in: brainly.com/question/2166533

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2 years ago

A dentist’s drill starts from rest. After 3.20 s of constant angu-lar acceleration, it turns at a rate of 2.51 3 104 re v/m i n.

Gekata [30.6K]

Answer:

$ΔTita = 4205.6 rad$

Explanation:

$w_{i}$ means initial angular velocity, which is 0 rev/min

$w_{f}$ means final angular velocity, which is $2.513*10^{4}rev/min$

t means time t= 3.20 s

one revolution is equivalent to 2πrad so the final angular velocity is:

$w_{f}$ = (2π/60) *2.513*10^{4} rad/s

$w_{f}$ = 2628.5 rad/s

so the angular acceleration, α will be:

α = 2628.5 rad/s / 3.20 s

$a = 821.40 rad/s^{2}$

so the rotational motion about a fixed axis is:

$w^{2} _{f} =w^{2} _{i}$ + 2αΔTita where ΔTita is the angle in radians

so now find the ΔTita the subject of the formula

ΔTita = $\frac{w^{2} _{f}-w^{2} _{i} }{2a}$

$ΔTita = ((2628.5)^{2} - (0 rev/min)^{2}) / 2* 821.40$

$ΔTita = 4205.6 rad$

7 0

3 years ago

Read 2 more answers

How long does it take for the velocity of the rain drop to reach 99% of its terminal velocity? (assume the conditions from part

vodomira [7]

If you think about it its part a and b

3 0

3 years ago

A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8

Umnica [9.8K]

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planet $m =6.75\times10^{24}\ kg$

Mass of star $m'=2.75\times10^{29}\ kg$

Radius of the orbit $r =8.05\times10^{7}\ km$

Using centripetal and gravitational force

The centripetal force is given by

$F = \dfrac{mv^2}{r}$

$F=m\omega^2r$

We know that,

$\omega=\dfrac{2\pi}{T}$

$F=m(\dfrac{2\pi}{T})^2r$ ....(I)

The gravitational force is given by

$F = \dfrac{mm'G}{r^2}$ ....(II)

From equation (I) and (II)

$m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}$

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

$T^2=\dfrac{4\pi^2R^3}{m'G}$

$T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}$

$T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}$

$T =3.34\times10^{7}\ s$

$T= 387.62\days$

Hence, The orbital period of the planet is 387.62 days.

4 0

3 years ago

Read 2 more answers