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monitta
4 years ago
5

At surface level, a diving bell has an air space of 4.25 m3. When used during an exploration, it is submerged 60.0 metres. At th

at point, what is the volume of the air space? Assume that the mean density of seawater is 1.025 g·cm−3, while the temperature is the same as on the surface.
Physics
1 answer:
Alinara [238K]4 years ago
8 0

Answer:

volume of the airspace=0.605m^3

Explanation:

Patm = 10^5 N/m^2

Depth= 60 metres

Pressure at the depth = ?

Density = 1.025 g/cm3 = 1025 kg/m3

P = Patm + hσg

P = 10^5 + 60*1025*9.8

P = 702700 N/m^2

P = 7.027 atm

Since the temperature is constant, Boyle’s law holds

P1V1 = P2V2

P1 = 1 atm  

P1, initial pressure of the bell(atm pressure) = 1 atm

The initial volume of the airspace, V1 = 4.25 m^3

Final pressure at the depth = 7.027 atm

Applying the Boyle’s lay

1*4.25 = 7.027 * V2

V2 = 4.25/7.027

V2 = 0.605m^3

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VMariaS [17]

Answer:

Explanation:

Given

final speed v = 3015ft/s

initial speed = 0m/s

Distance S = 4146m/s

Required

Time

Using the equation of motion to get the time;

S = (v+u)/2 * t

since 1m = 3.28084ft'

4146m = 3.28084 * 4146 = 13,602.36264feet

13,602.36264 = 3015+0/2 * t

13,602.36264 = 1,507.5t

t = 13,602.36264/ 1,507.5

t = 9.023 secs

Hence it will take 9.023 seconds the rescuer die the biller to strike a target

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3 years ago
The maximum current output of a 60 ω circuit is 11 A. What is the rms voltage of the circuit?
Scilla [17]

Answer:

sorry

Explanation:

I dont know the anwser to that one

6 0
3 years ago
A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m
Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

  • OA = range of initial position of the airplane from the point of observation = 375 m
  • OB = range of the final position of the airplane from the point of observation = 797 m
  • \theta = angle of the initial position vector from the observation point = 43^\circ
  • \alpha = angle of the final position vector from the observation point = 123^\circ
  • \vec{AB} = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} =  (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} =  (-708.34\ \hat{i}+412.67\ \hat{j})\ m

The magnitude of the displacement vector = \sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0
3 years ago
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Tanzania [10]
The energy of one massive object has in relation with another massive object due to gravity is called gravity
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a boy looks at the reflection of his digital watch in a plane mirror and thinks the time is 10:11. what is the correct time?
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Answer:

11:10 will be the time. reflection causes the object to be flipped when you see its image at the mirror

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