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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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To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

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\lambda = Wavelength \rightarrow 35*10^{-2}m

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\omega = 2\pi *57.14

\omega = 359.03rad/s

Now the maximum speed from the simple harmonic movement is given by

V_{max} = A\omega

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3 years ago
Read the paragraph below and answer the question that follows:
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B. Helium because it is constantly being made in the sun

Explanation:

From the given paragraph, we can conclude that helium is the most common element in a star such as the sun because it is constantly being made.

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A fan cart initially has an acceleration of 1.6m/s/s when it's fan is directed straight backwards. If you rotate the fan by 45°,
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Answer:

A Fan Cart Initially Has An Acceleration Of 1.6m/s/s When It's Fan Is Directed Straight Backwards. If You Rotate The Fan By 45o, By What Percentage Do You Expect The Fan Cart's Thrust To Decrease? (Answer Should Be In Units Of 96)

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b. 29%

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d. 50%

The correct answer is d.

d. 50%

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or 0.707×D. The thrust becomes 0.25×π×(0.707×D)²×ρ×v×Δv

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A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter
kifflom [539]

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b). 1.78 \frac{m}{s}

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b).

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V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}

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This galaxy belongs to the elliptical galaxy category. This is because it does not have spiral arms.

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