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4 years ago

The LR5 is the specialist submarine for underwater rescue. The average density of seawater is 1028 kg/ m3.

Physics

1 answer:

sladkih [1.3K]4 years ago

5 0

Answer:

P = 7196 [kPa]

Explanation:

We can solve this problem using the expression that defines the pressure depending on the height of water column.

P = dens*g*h

where:

dens = 1028 [kg/m^3]

g = 10 [m/s^2]

h = 700 [m]

Therefore:

P = 1028*10*700

P = 7196000 [Pa]

P = 7196 [kPa]

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Soft ball,kick ball,wiffle ball??

Anvisha [2.4K]

1. No they aren’t because they all belong to different sports and are used differently

6 0

3 years ago

Read 2 more answers

uppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviation of 0.8 ou

Brrunno [24]

Answer:

x= 9.53 ounces

Explanation:

Given that

Mean ,μ= 9 ounces

Standard deviation ,σ=0.8 ounces

He wants to sell only those potatoes that are among the heaviest 25%.

P=25% = 0.25

When P= 0.25 then Z=0.674

Lest take x is the the minimum weight required to be brought to the farmer's market.

We know that

x = Z . σ + μ

x= 0.674 ₓ 0.8 + 9 ounces

x= 9.53 ounces

3 0

3 years ago

A ball is thrown straight up into the air and passes a window 3 seconds after being released. It passes the same window on its w

melomori [17]

The initial velocity of the ball is 55.125 m/s.

<h3>Initial velocity of the ball</h3>

The initial velocity of the ball is calculated as follows;

During upward motion

h = vi - ¹/₂gt²

h = vi - 0.5(9.8)(3²)

h = vi - 44.1 ----------------- (1)

During downward motion

h = vi + ¹/₂gt²

h = 0 + 0.5(9.8)(1.5)²

h = 11.025 ----------- (2)

solve (1) and (2) together, to determine the initial velocity of the ball

11.025 = vi - 44.1

vi = 11.025 + 44.1

vi = 55.125 m/s

Thus, the initial velocity of the ball is 55.125 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

#SPJ1

8 0

1 year ago

Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0

3 years ago

Convert mm3 into m3.

bixtya [17]

Answer:

$1 \times 10 { - }^{9}$

cubic metre or 1e-9

Explanation:

•By division. Number of cubic millimetre divided(/) by 1000000000, equal(=): Number of cubic metre.

•By multiplication. 83 mm3(s) * 1.0E-9 = 8.3E-8 m3(s)

4 0

3 years ago