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

I tried the experiment of dropping candy into a liter of soda and caused a big explosion of foamy soda. Is that a physical or Ch

Aleks04 [339]

Answer:

physical change

Explanation:

its a physical change because its still soda it hasn't changed into anything else than what it already is

6 0

3 years ago

A particle leaves the origin with a speed of 2.1 times 106 m/s at 30 degrees to the positive x axis. It moves in a uniform elect

Verizon [17]

Answer:

-1449.69404 N/C

Explanation:

u = Velocity of particle = $2.1\times 10^6\ m/s$

$\theta$ = Angle = 30°

x = Distance = 1.5 cm

m = Mass of electron = $9.11\times 10^{-31}\ kg$

q = Charge of electron = $-1.6\times 10^{-19}\ C$

In the case of projectile motion

$x=utcosA\\\Rightarrow t=\dfrac{x}{ucosA}$

The force of on the particle will balance the Electric force

$ma=qE\\\Rightarrow a=\dfrac{qE}{m}$

Now

$y=utsin\theta-\dfrac{1}{2}at^2\\\Rightarrow y=utsin\theta-\dfrac{1}{2}\dfrac{qE}{m}t^2$

If y = 0

$0=utsin\theta-\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow utsin\theta=\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow t=\dfrac{2musin\theta}{qE}$

$\dfrac{x}{ucosA}=\dfrac{2musin\theta}{qE}\\\Rightarrow E=\dfrac{2mu^2sin\theta cos\theta}{xq}\\\Rightarrow E=\dfrac{2\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\times sin30\times cos30}{1.5\times 10^{-2}\times (-1.6\times 10^{-19})}\\\Rightarrow E=-1449.69404\ N/C$

The electric field is -1449.69404 N/C

8 0

3 years ago

Hydraulic systems utilize Pascal's principle by transmitting pressure from one cylinder (called the primary) to another (called

Paul [167]

Answer:

F1= 122.93 N

Explanation:

Pascal´s Principle

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

P=F/A

P1=P2

$\frac{F_{1} }{A_{1}} = \frac{F_{2} }{A_{2}}$ Formula (1)