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sashaice [31]
3 years ago
13

Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air

Force's Aero Med lab, pioneering research into the accelerations which humans could tolerate and the types of physiological effects which would result. After several runs with a 185-pound dummy named Oscar Eightball, Captain Stapp decided that tests should be conducted upon humans. Demonstrating his valor and commitment to the cause, Stapp volunteered to be the main subject of subsequent testing. Manning the rocket sled on the famed Gee Whiz track, Stapp tested acceleration and deceleration rates in both the forward-sitting and backward-sitting positions. He would accelerate to aircraft speeds along the 1200-foot track and abruptly decelerate under the influence of a hydraulic braking system. On one of his most intense runs, his sled decelerated from 282 m/s (632 mi/hr) to a stop at -201 m/s/s. Determine the stopping distance and the stopping time.
Physics
2 answers:
Gemiola [76]3 years ago
6 0

Answer:Stopping Distance= 197.82 m

             Stopping time= 1.403 s

Explanation:

<u>Given that</u>:

  • initial velocity, u=282 ms^{-1}
  • acceleration, a= -201 ms^{-2}
  • final velocity, v= 0 ms^{-1} ∵ the body comes to the rest finally

<u>To find</u>:

  • Stopping time, t
  • stopping distance, s

From the given and asked data we identify the required equations of motion.

For calculating the stopping time: v=u+at ⇒ t=\frac{v-u}{t}

t=\frac{0-282}{-201} = 1.403 s

For calculating the stopping distance:

v^{2} =u^{2} +2as ⇒ s= \frac{v^{2} -u^{2} }{2a}

<em>putting the respective values</em>

s= \frac{0^{2}-282^{2}}{2(-201)} = 197.82 m

valentina_108 [34]3 years ago
4 0

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: v_f=v_0+a*t, where v_f is the final velocity, v_0 is the initial velocity, a the acceleration, and t is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s}  }{-201\frac{m}{s^2} }=1.40s, where v_f=0\frac{m}{s} because the sled is totally stopped, v_0=282\frac{m}{s} is the velocity of the sled before braking and, a=-201\frac{m}{s^2} is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration:x_f-x_0=v_0*t+\frac{1}{2}*a*t^2, where x_f-x_0 is the distance traveled, v_0 is the initial velocity, t the time of the process and, a is the acceleration of the process.

Then for this case the relationship becomes: x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m.

<u>Note that the acceleration is negative because is a braking process.</u>

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Which two structures are not found in animal cells?
Alinara [238K]

Answer:

C

Explanation:

So, lets start off by thinking about what we know is in both cells, and rule out from there.

There are just some basic things you need in eukaryotic cells, those include the following:

  • Nucleus
  • cell membrane
  • mitochondria
  • cytoplasm(dont get confused with chloroplast)
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  • Golgi vescile
  • Golci apparatus
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Here are jsut a few examples of what can be found in both plant and animal cells, however we can already rule out all of the following except for:

C

Because remember, c talks about the chlorplast, which is not the chlorplasm found in both plant anf animal cells, its just the chloroplast found in plant cells.

Although I gave the asnwer to you. Here are some things that are in plant cells that arent in animal cells:

  • Cell wall
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Animal cells on the other hand, have a few things that plant cells do not as well:

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Hope this helps! ;)

7 0
2 years ago
The burning of fossil fuels releases gas into the air, the gas is called?
NemiM [27]

Answer:

They mainly consist of carbon dioxide and carbon mono oxide gas

I hope this helps

8 0
3 years ago
A wave travels a distance of 60cm in 3s. The distance blw successive crests of thd wave is 4cm. What is the frequency?
schepotkina [342]
Data:
ΔS = 60cm
Δt = 3s
Vm = ?

Vm = ΔS/Δt 
Vm =  \frac{60}{3}
Vm = 20cm/s

We have:

Vm = v
ΔS = λ
Δt = T

v =  \frac{\lambda}{T}
20 = \frac{\lambda}{4}
\lambda = 20*4
\lambda = 80cm

Soon:

v = \lambda*f
20 = 80*f
80f = 20
f =  \frac{20}{80} simplify( \frac{\div20}{\div20} )= \frac{1}{4}
\boxed{f = 0,25Hz}


6 0
3 years ago
An elastic conducting material is stretched into a circular loop of 11.2 cm radius. It is placed with its plane perpendicular to
anygoal [31]

Answer:

0.426 volts

Explanation:

It is given that,

The radius of a circular loop, r = 11.2 cm = 0.112 m

An elastic conducting material is stretched into a circular loop.

It is placed with its plane perpendicular to a uniform 0.880 T magnetic field.

The radius of the loop starts to shrink at an instantaneous rate of 68.8 cm/s, dr/dt = 0.688 m/s

We need to find the emf induced in the loop at that instant.

\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{d}{dt}(BA)\\\\=\dfrac{d}{dt}(\pi r^2 B)\\\\=\pi B\dfrac{d}{dt}(r^2)\\\\=2\pi B r\dfrac{dr}{dt}\\\\=2\pi \times 0.88\times 0.112\times 0.688\\\\=0.426\ V

So, the magnitude of induced emf is 0.426 volts.

4 0
2 years ago
On the average, about what percentage of the solar energy that strikes the outer atmosphere eventually reaches the earth's surfa
almond37 [142]
I think the answer to your question is twenty-four percent.
5 0
3 years ago
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