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3 years ago

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A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 17 feet. The ball is started in moti

on from the equilibrium position with a downward velocity of 4 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) . Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second.

2 answers:

valkas [14]3 years ago

7 0

Answer:

y = (0 - 4 * sqrt(6)/48) * exp ((-16 -4 * sqrt(6) )*t + (0 + 4 * sqrt (6) /48) * exp ((-16 + 4 * sqrt (6)) * t)

Explanation:

Assume that the gravitational acceleration is 32 feet per second square.
y = (0 - 4 * sqrt(6)/48) * exp ((-16 -4 * sqrt(6) )*t + (0 + 4 * sqrt (6) /48) * exp ((-16 + 4 * sqrt (6)) * t)

Romashka-Z-Leto [24]3 years ago

5 0

twelve because it is the answer

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3.12 ** To illustrate the use of a multistage rocket consider the following: (a) A certain rocket carries 60% of its initial mas

madreJ [45]

Answer:

$0.916v_{ex}$

Explanation:

A multistage rock is also known as a step rocket and it is a form of vehicle that can use more than a rocket stage. The rocket stage typically contains propellant as well as its engine. The final speed of the rocket can be estimated using the equation below:

$v = v_{ex} ln(\frac{m_{o}}{0.4m_{o}})$ $= v_{ex}*ln(2.5) = v_{ex}*0.916 = 0.916v_{ex}$

Therefore, the maximum speed of the rocket is $0.916v_{ex}$

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4 years ago

A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction

finlep [7]

consider the motion in x-direction

$v_{ox}$ = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

$a_{x}$ = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = $v_{ox}$ t + (0.5) $a_{x}$ t²

100 = $v_{ox}$ (4.60)

$v_{ox}$ = 21.7 m/s

consider the motion along y-direction

$v_{oy}$ = initial velocity in y-direction = ?

Y = vertical displacement = 0 m

$a_{y}$ = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = $v_{oy}$ t + (0.5) $a_{y}$ t²

0 = $v_{oy}$ (4.60) + (0.5) (- 9.8) (4.60)²

$v_{oy}$ = 22.54 m/s

initial velocity is given as

$v_{o}$ = sqrt(( $v_{ox}$ )² + ( $v_{oy}$ )²)

$v_{o}$ = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

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4 years ago

The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in 1851. Consider a l

Ket [755]

It is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

Given ,

The motion of a transparent medium influences the speed of light .

The water moves with speed v in a horizontal pipe .

Assume that the light travels in the same direction as the water moves .

The speed of the light with respect to the water is c/n

Where n = 1.33 is the refractive index of water .

Let us assume ,

u' be the speed of light in water , in the frame moving with the water .

u' is related to the refractive index of water ,n as :

u'=c/n

where , c is the speed of light .

let , u be the speed of light in water in the lab frame .

Now , u and u' are related as : u = (u'+ v )/(1+ u'v/c^2)

Here v is the speed of water in the horizontal pipe .

we know the value of u' , so by substituting the value , we will get ,

u= (c/n+ v)/(1+cv/nc^2)

u= c/n(1+ nv/c)/(1+v/nc)

(b) We have , v<<c

v/c<<1 .

so , (1+v/nc )^-1 = (1-v/nc)

Now substituting this , we will get ,

u = c/n(1+nv/c) (1-v/nc)

u≈c/n(1+ nv/c-v/cn)

u≈c/n + v - v/n^2

Hence , it is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

Learn more about speed here :

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Disclaimer : incomplete question , here is the complete question .

Question: The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n=1.33 is the index of refraction of water.(a) Use the velocity transformation equation to show that the speed of the light measured in the laboratory frame isu = c/n (1 + nv/c / 1+ v/nc) . (b) show that for v<<c , the expression from part (a) becomes , to a good approximation , u ≈ c/n + v - v/n^2 .

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2 years ago

Are you aware of human rights violation happening in the community and explain

mariarad [96]

Answer:

Individuals who commit serious violations of international human rights or humanitarian law, including crimes against humanity and war crimes, may be prosecuted by their own country or by other countries exercising what is known as “universal jurisdiction.”

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3 years ago

Two loudspeakers are placed on a wall 2 m apart. A listener stands directly in front of one of the speakers, 81.7 m from the wal

OverLord2011 [107]

Answer:

The phase difference is $\Delta \phi = 1.9995 rad$

Explanation:

From the question we are told that

The distance between the loudspeakers is $d = 2m$

The distance of the listener from the wall $D = 81.7 \ m$

The frequency of the loudspeakers is $f = 4450Hz$

The velocity of sound is $v_s = 343 m/s$

The path difference of the sound wave that is getting to the listener is mathematically represented as

$\Delta z =\sqrt{d^2 + D^2} -D$

Substituting values

$\Delta z =\sqrt{2^2 + 81.7^2 } -81.7$

$\Delta z =0.0245m$

The phase difference is mathematically represented as

$\Delta \phi$ = $\frac{2 \pi}{\lambda } * \Delta z$

Where $\lambda$ is the wavelength which is mathematically represented as

$\lambda = \frac{v_s }{f}$

substituting value

$\lambda = \frac{343 }{4450}$

$\lambda = 0.0770 m$

Substituting value into the equation for phase difference

$\Delta \phi$ = $\frac{2 * 3.142 * 0.0245}{0.0770}$

$\Delta \phi = 1.9995 rad$

8 0

4 years ago