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3 years ago

15

During a field experiment about speed, a scientist created the chart above. The chart shows distance and time measurements for a

racing car on the straight section of a race track. What is the racing car's speed?

1 answer:

BaLLatris [955]3 years ago

6 0

Answer:

yes

Explanation:

its not a good thing for the rest of your life but you have a PS4

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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

Something that has many particles in a small space would have a ————- density

Diano4ka-milaya [45]

Answer:

a high density i believe

Explanation:

4 0

3 years ago

Read 2 more answers

What happens to the wavelength of a wave if you increase the frequency?

dezoksy [38]

The product of (wavelength) times (frequency) is always the same number ...
the speed of the wave in whatever material it's traveling through. So if the
frequency is increased, then the wavelength must <em><u>de</u></em>crease by the same
factor, in order to keep the product the same.

4 0

3 years ago

) Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of ene

Angelina_Jolie [31]

Answer:

8100W

Explanation:

Let g = 10m/s2

As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

$P = \dot{E} = \dot{m}gh = 15*10*60 = 9000 J/s$ or 9000W

Since 10% of this is lost to friction, we take the remaining 90 %

P = 9000*90% = 8100 W

3 0

3 years ago

Two poles are connected by a wire that is also connected to the ground. The first pole is 20ft tall and the second pole is 10ft

Natasha2012 [34]

Answer: 31.6ft

Explanation:

Check the attachment for the diagram.

According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|

that is 20ft - 10ft = 10ft

According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.

|AE|^2 + |EC|^2 = |AC|^2

10^2 + 30^2 = |AC|^2

100 + 900 = |AC|^2

|AC| = √1000

|AC| = 31.6ft

Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.

6 0

4 years ago