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2 years ago

7

For many years Colonel John P. Stapp, USAF, held the world's land speed record. He participated in studying whether a jet pilot

could survive emergency ejection. On March 19, 1954, he rode a rocket-propelled sled that moved down a track at a speed of 632 mi/h. He and the sled were safely brought to rest in 1.40s
(a) Determine the negative acceleration he experienced.
__ m/s2
(b) Determine the distance he traveled during this negative acceleration.
__ m.

1 answer:

Margarita [4]2 years ago

3 0

Answer:

(a) -202 m/s²

(b) 198 m

Explanation:

Given data

Initial speed (v₀): 283 m/s

$\frac{632mi}{h} .\frac{1609.34m}{1mi} .\frac{1h}{3600s} =283m/s$

Final speed (vf): 0 (rest)

Time (t): 1.40 s

(a) The acceleration (a) is the change in the speed over the time elapsed.

a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²

(b) We can find the distance traveled (d) using the following kinematic expression.

y = v₀ × t + 1/2 × a × t²

y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²

y = 198 m

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How fast does a 2 MeV fission neutron travel through a reactor core?

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Answer:

The answer is $1.956 \times 10^7\ m/s$

Explanation:

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From the last equation,

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where

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2 years ago

Easy defimation of newtons second law of motion

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Hi pupil here's your answer ::

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3 years ago

A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo

Svetllana [295]

Answer:

The gravitational potential energy of the nickel at the top of the monument is 8.29 J.

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We can find the gravitational potential energy using the following formula.

$GPE=mgh$

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The nickel has a mass $m=0.005 \,kg$ , and it is a the top of Washington Monument.

The Washington Monument has a height of $h=555 \, ft$ , thus we need to find the equivalence in meters using unit conversion in order to find the gravitational potential energy.

Converting from feet to meters.

Using the conversion factor 1 m = 3.28 ft, we have

$h = 555 \, ft \times \cfrac{1 \, m}{3.28 \, ft}$

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Finding Gravitational Potential Energy.

We can replace the height and mass on the formula

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2 years ago

Read 2 more answers

An object is falling from a height of 7.5 meters. At what height will it’s velocity be 7 meters/second?

Nady [450]

One of the equations of gravity is this:
${v}^{2} = {u}^{2} + 2gh$
Where v = final velocity which is 7m/s
u = initial velocity which is 0 for objects falling from a height
g = acceleration due to gravity and it is approximately 10m/s^2. It's a constant so pretty much remember this number. It's positive since the work being done is caused by gravity (in other words, it's falling down). It can also be negative if the work being down is against gravity (in other words, it's going up)
h = height of object

Substitute for the values and you should have something like this
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$h = \frac{49}{20}$
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2 years ago

A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

$\lambda = \frac{2\pi}{k}$

$\lambda = (frac{2\pi}{\frac{\pi}{11})
$

$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

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3 years ago