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Angelina_Jolie [31]

3 years ago

15

The Southwest Indian Ridge, shown in red, moves at an average rate of 20 mm/year, making it among the ultraslow spreading ridges

" on Earth. In how many years will the ridge move 100 mm?

2 answers:

blagie [28]3 years ago

7 0

v = average speed of movement of the Southwest Indian Ridge = 20 mm/year

d = distance moved by the Southwest Indian Ridge = 100 mm

t = number of years required to move distance "d"

distance traveled is given as

d = v t

inserting the above values in the formula

100 mm = (20 mm/year) t

dividing both side by 20 mm/year

t = 100 mm/(20 mm/year)

t = 5 years

slega [8]3 years ago

4 0

Answer:5 years

Explanation:

I just took the test.

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A 2300 kg sailboat is moving west at 5.5 m/s when an eastward wind

pogonyaev

The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of

950 N = (2300 kg) <em>a</em>

<em>a</em> = (950 N) / (2300 kg)

<em>a</em> ≈ 0.413 m/s²

Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of

<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)

<em>v</em> = -0.75 m/s

which means the wind slows the boat down to a velocity of 0.75 m/s westward.

5 0

3 years ago

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

<u>Velocity just before opening the parachute:</u>

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

<u>Now the height fallen in the remaining time using parachute:</u>

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

3 years ago

Please help me I am desperate this is due now

elena55 [62]

Answer:

Distance travelled is 7 meters and the displacement is 3 meters

7 0

3 years ago

Read 2 more answers

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

3 years ago

Which of the following elements do living things have that volcanic rocks do not have?

jek_recluse [69]

Answer: silicon,or maybe none.

Explanation: I searched it up not really sure sorry

8 0

3 years ago