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3 years ago

Name and describe the three types of binary systems.

Physics

1 answer:

Pachacha [2.7K]3 years ago

5 0

Answer and Explanation:

The binary system is a star system about half of the star are the part of the binary system

The important binary systems are

Visual binary : in visual binary system the two stars resolved visually with nay sort of optical device the orbit of binary orbit is very large about center of mass.
Spectroscopic binary : in this binary system the stars are detected by close analysis of light and after detection it is found that there are two seller spectrum present instead of one.
Eclipsing binary : eclipsing binary system is a several types of variable stars this binary system is very important in astrophysics

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Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

3 years ago

Moustapha jones drives east at 100km/hr for 3 hours then back west at 80km/hr for 1.5 hours. which pair of answers gives his ave

olya-2409 [2.1K]

The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.

The total distance cover in east direction is=100*3=300 km

The total distance cover in the west direction=80*1.5=120 km

The total distance covered is =300+120=420 km

And Total displacement of the car is =300-120=180 km

As we know that the average speed is given as

Avg Speed =Total Distance / Total time

=420/4.5=93.33 km/hr