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Soloha48 [4]
3 years ago
8

An exploration submarine should be able to descend 1200 m down in the ocean. If the ocean density is 1020 kg/m3, what is the max

imum pressure on the submarine hull?
Physics
1 answer:
motikmotik3 years ago
8 0

Answer:

11995200 N/m²

Explanation:

Pressure: The is the ratio of force to the surface area in contact. The S.I unit of pressure is N/m².

Generally pressure in fluid can be expressed as

P = ρgh......................... Equation 1

Where P = maximum pressure on the submarine hull, ρ = Density of ocean, h = depth of ocean, g = acceleration due to gravity.

Given: h = 1200 m, ρ = 1020 kg/m³

Constant: g = 9.8 m/s²

Substitute into equation 1

P = 1200(1020)(9.8)

P = 11995200 N/m²

Hence the maximum pressure on the submarine hull = 11995200 N/m²

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This is physics 11th grade and a homework question I don’t understand how to do this or what the question is asking me
Alexxx [7]

a) Frequency is the number of complete oscillations per second. Looking at the graph, there are 9 complete oscillations in 5 seconds. Thus,

Frequency = 9/5 = 1.8 oscillations per second

Frequency = 1.8 Hz

Period = 1/frequency = 1/1.8

Period = 0.056 s

b) When we differenctiate displacement with respect to time, the result is velocity.

Recall, period = 1/f = 5/9 cycles

1/4 cycle behind = 1/4 x 5/9 = 5/36

It is delayed with 5/36 sec with respect to displacement.

5/36 sec = 0.139 sec

Acceleration = first derivative of velocity = second derivative of displacement = 1/4 cycle behind velocity = 1/2 cycle behind displacement =

5/36 = 0.139 sec delayed with respect to velocity

= 5/18 = 0.2777 secs delayed with respect to displacement

Thus, the number of seconds out of phase with the displacements is 0.278 seconds

c) The formula for calculating the period of an ideal pendulum anywhere is

T = 2π√length/local gravity). We would calculate the local gravity.

From the information given,

length = 0.2

T = P = 5/9

Thus,

5/9 = 2π√0.2/local gravity)

(5/9)/2π = √0.2/local gravity

Square both sides. It becomes

[(5/9)/2π]^2 = 0.2/local gravity

local gravity = 0.2/[(5/9)/2π]^2

local gravity = 25.56 m/s^2

Thus,

acceleration due to gravity = 25.56 m/s^2

Recall, earth's gravity = 9.8 m/s^2

number of g forces = 25.56/9.8

number of g forces = 2.61

6 0
1 year ago
Flapping flight is very energy intensive. A wind tunnel test
steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight=\frac{P}{m}

Using the formula

Metabolic power for starting flight=\frac{12}{0.089}

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

4 0
3 years ago
Some birds migrate 20,000 miles. If 1 mile equals 1.6 kilometers, calculate the distance these birds fly in kilometers.
faust18 [17]
Set up a proportion

1 mile/1.6 km = 20,000miles/x

Cross Multiply
x = (20,000) * (1.6)
x = 32,000 kilometers
6 0
3 years ago
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schepotkina [342]

Answer:

thanks for the points

Explanation:

8 0
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