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3 years ago

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She falls out of a plane 2 miles up, equipped with a parachute as well as a backup parachute, they both fail. How much time does

it take for her to hit the surface?

1 answer:

Minchanka [31]3 years ago

8 0

D=Vot+1/2at^2

Vot= 0 since there is no initial velocity in the y direction.

a=gravitational acceleration which is - 32 ft/s^2 in imperial units I believe.

She travels a distance of -2 miles, so

-2 miles= 1/2(-32ft/s^2)t^2

t= square root((- 4 miles)/(-32ft/s^2))

I don't know imperial units, but convert miles to feet then solve for t.

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How much time passes when an object travels at a constant speed of 17 m/s over a distance of 323 m? Identify Variables, write fo

k0ka [10]

Answer: 19 s

Explanation:

The relationship between speed, distance and time is given by:

velocity= distance/time

time= distance/velocity= 323 m/17 m/s= 19 s

8 0

3 years ago

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:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

d)

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:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

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.

8 0

3 years ago

What happens during destructive interference?

Masteriza [31]

They will subtract to form a combined wave with a lower amplitude

3 0

3 years ago

How would the attractive force between two spheres change if the mass of one sphere was doubled?

Dominik [7]

The force of attraction between two objects can be illustrated using Newton's Law of Universal Gravitation.
The relation between the different parameters is shown in the attached image.

Now, from the relation, we can deduce that the force between the two objects is directly proportional to the masses of the two objects.

This means that, if the mass of one object is doubled, then the force between the two objects will also be doubled.

8 0

3 years ago

A year 11 pupil with a mass of 55kg swinging back on their chair and falling off it at a speed of 0.6m/s. What is his kinetic en

posledela

Answer:

Uk = 9.9 J

Explanation:

To calculate the kinetic energie (Uk), you can make use of this formula:

Uk = 0.5 * m * v²

given m = 55 kg and v = 0.6 m/s

Substituting in the formula gives:

Uk = 0.5 * 55 * (0.6)²

Uk = 0.5 * 55 * 0.36

Uk = 9.9 J

Extra:

Now let's examine the formula in relation to the SI units. <em>If you understand the following, it will give you great insight in how smart Phisics is inter twained by looking at formulas and their standard units. It will save you time in future to convert formulas, if you use the right standard units.</em>

The formula for kinetic energie is:

Uk = 0.5 * m * v²

Standard SI unit for mass m is kg.

Standard SI unit for speed v is m/s.

So v * v = v² and therefore v² must have the standard SI unit of m²/s².

From the formula, you see that the unit of Uk must be kg*m²/s² and since Uk is normally given in J, these both forms must be the same !

The main unit for Uk is the Joule. <em>Now</em><em> </em><em>please</em><em> </em><em>see</em><em> </em><em>the</em><em> </em><em>picture</em><em>,</em><em> </em><em>which</em><em> </em><em>shows</em><em> </em><em>the </em><em>relation</em><em> </em><em>between </em><em>the </em><em>J </em><em>and </em><em>other</em><em> SI units</em><em>.</em><em> </em><em>Please</em><em> </em><em>understand</em><em> </em><em>that</em><em> </em><em>you</em><em> </em><em>can</em><em> </em><em>construct</em><em> </em><em>your</em><em> </em><em>'own'</em><em> </em><em>formulas</em><em> </em><em>based</em><em> </em><em>these</em><em> </em><em>units</em><em>.</em><em> </em><em>Now</em><em> </em><em>here</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>time</em><em> </em><em>saver</em><em>:</em>

Because almost always the right units are <em>given</em> in a question, or because sometimes you can look up a constant in a table with an exotic and seemingly complicated unit, but that says a lot about the formula which must have been some how involved!

<em>By this, I hope you now understand the implication of using the right standard SI units and how that can help you figure out what formula is needed.</em>

3 0

3 years ago