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Soloha48 [4]
2 years ago
6

Professional application dr. john paul stapp was u.s. air force officer who studied the effects of extreme deceleration on the h

uman body. on december 10, 1954, stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! calculate his (a) acceleration and (b) deceleration. express each in multiples of ? (9.80 m/s2) by taking its ratio to the acceleration of gravity.
Physics
1 answer:
aleksley [76]2 years ago
5 0
A) Acceleration= (282m/s-0m/s)/5.0s= 56.4m/s^2
b) Deceleration=(0m/s-282m/s)/1.40s=201.4m/s^2 (here we reject negative because deceleration already take into account negative)
c) for the first one the multiple will be 5.76g
for the second one it will be 20.6g
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Leokris [45]

By Newton's second law,

• the net force acting vertically on the crate is 0, and

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0   ==>   <em>n</em> = <em>mg</em> = 1470 N

where <em>n</em> is the magnitude of the normal force; and

• the net force acting in the horizontal direction on the crate is also 0, with

∑ <em>F</em> = <em>f</em> - <em>b</em> = 0   ==>   <em>b</em> = <em>f</em> = <em>µn</em> = 0.645 (1470 N) = 948.15 N

where <em>b</em> is the magnitude of the braking force, <em>f</em> is (the maximum) static friction, and <em>µ</em> is the coefficient of static friction. This is to say that static friction has a maximum magnitude of 948.15 N. If the brakes apply a larger force than this, then the crate will begin to slide.

Note that we are taking the direction of the truck's motion as it slows down to be the positive horizontal direction. The brakes apply a force in the negative direction to slow down the truck-crate system, and static friction keeps the crate from sliding off the truck bed so that the frictional force points in the positive direction.

Let <em>a</em> be the acceleration felt by the crate due to either the brakes or friction. Use Newton's second law again to solve for <em>a</em> :

<em>f</em> = <em>ma</em>   ==>   <em>a</em> = (948.15 N) / (150.0 kg) = 6.321 m/s²

With this acceleration, the truck will come to a stop after time <em>t</em> such that

0 = 50.0 km/h - (6.321 m/s²) <em>t</em>   ==>   <em>t</em> ≈ (13.9 m/s) / (6.321 m/s²) ≈ 2.197 s

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8 0
2 years ago
A ball is thrown straight up at 20 m. What is the balls velocity as it hits the ground?
deff fn [24]

Answer:

<em>The velocity of the ball as it hit the ground = 19.799 m/s</em>

Explanation:

Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.

v² = u² + 2gs.......................... Equation 1

Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball

<em>Given: s = 20 m, u = 0 m/s</em>

<em>Constant: g = 9.8 m/s²</em>

<em>Substituting these values into equation 1,</em>

<em>v² = 0 + 2×9.8×20</em>

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<em>Therefore the velocity of the ball as it hit the ground = 19.799 m/s</em>

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A disk with a radius of R is oriented with its normal unit vector at an angle Θ with respect to a uniform electric field. Which
marissa [1.9K]

Answer:

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⇒ Ф = E·A·e×n = E·A·cos(angle(e,n)) = E·A·cos(Θ)

Since in this case the area is for a disk of radius R, A=\pi R^{2}

So, Ф = E·πR^2·cos(Θ)

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