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dexar [7]
3 years ago
13

A 25-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, ho

w fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 20 ft from the wall? Answer with 2 decimal places.
Physics
1 answer:
lyudmila [28]3 years ago
3 0

Let x be the distance between the base of the ladder and the bottom of the wall, and y the distance between the top of the ladder and the bottom of the wall, so that

x^2+y^2=(25\,\mathrm{ft})^2

Differentiate both sides with respect to time t:

2x\dfrac{\mathrm dx}{\mathrm dt}+2y\dfrac{\mathrm dy}{\mathrm dt}=0

When x=20\,\rm ft, the top of the ladder is

y=\sqrt{(25\,\mathrm{ft})^2-(20\,\mathrm{ft})^2}=15\,\mathrm{ft}

above the ground. Then, given that the bottom of the ladder slides away from the wall at a rate of \dfrac{\mathrm dx}{\mathrm dt}=0.18\dfrac{\rm ft}{\rm s}, we have

2(20\,\mathrm{ft})\left(0.18\dfrac{\rm ft}{\rm s}\right)+2(15\,\mathrm{ft})\dfrac{\mathrm dy}{\mathrm dt}=0\implies\dfrac{\mathrm dy}{\mathrm dt}=-0.24\dfrac{\rm ft}{\rm s}

That is, the top of the ladder is sliding downward at a rate of 0.24 ft/s.

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Explanation:

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6 0
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A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be
lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × 10^{-3} m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36  

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ)       ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v  /  2     ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × 10^{-3})²×ω²× 34.64 )   /  2

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A helium atom (mass 4.0 u) moving at 598 m/s to the right collides with an oxygen molecule (mass 32 u) moving in the same direct
labwork [276]

Answer:

Speed of the helium after collision = 246 m/s

Explanation:

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Given that initially both are moving in the same direction and lets take they are moving in the right direction.

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There is no any external force on the masses that is why the linear momentum will be conserve.

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598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445

v₁ = 246 m/s

Speed of the helium after collision = 246 m/s

6 0
3 years ago
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