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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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Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

Explanation :

Given that,

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Final pressure = 1.0 atm

We need to calculate the initial volume

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P_{1}V_{1}=nRT

V_{1}=\dfrac{nRT}{P_{1}}

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(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}

Put the value into the formula

T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}

T_{2}=233.7\ K

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}

Put the value into the formula

V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}

V_{2}=0.00786\ m^3

V_{2}=7.86\times10^{-3}\ m^3

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}

V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}

V_{2}=0.0057\ m^3

V_{2}=5.7\times10^{-3}\ m^3

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

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