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marysya [2.9K]
3 years ago
5

a rescue line is to be thrown horizontally from the bridge of a cruise ship at 40.0 m above sea level to a life boat 50.0 m away

with what horizontal speed should the rescue line be thrown?
Physics
1 answer:
Gemiola [76]3 years ago
8 0
There is a 40m drop from the bridge, so we can work out the time of the vertical component first (how long it will take to reach the water) 

do this by using a SUVAT equation s=ut+0.5at^2
this rearranges to t= sqrt(2s/a)
- note that u has been removed since there is no initial velocity.
 
sqrt((2*40)/9.81)= 2.86 seconds 

this means that the horizontal component of 50m has so be covered in 2.86s 
and speed = distance/ time 

50m/2.86s=17.48ms^-1

the line has to be thrown at 17.48ms^-1 or more to reach the life boat. 
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f_{c}=\dfrac{1}{2\pi R C}

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f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}

f_{c}=1198.45\ Hz

(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})

Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=2.280\ Volt

(C). We need to calculate the V_{R} when f = f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=3.606\ Volt

(D). We need to calculate the V_{R} when f = 2f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

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8 0
4 years ago
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fenix001 [56]

Answer:

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3 years ago
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