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3 years ago

12

Payment to beneficiaries who were named by the insured person

1 answer:

Zanzabum3 years ago

4 0

Death benefit from a Life insurance policy

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B. to achieve sustainable development

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Tech B says that long-term fuel trims that are positive means that the PCM is leaning out the fuel mixture from the base pulse-w

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Explanation:

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What are the coordinates of the centroid of this figure?

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2 years ago

Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph

kicyunya [14]

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

$S.G=\frac{\rho_{sw}}{\rho_{w}}$

Here, density of sea water is $\rho_{sw}$ , surface gravity is S.G and density of water is $\rho_{w}$ .

Substitute all the values in the above equation as follows:

$S.G=\frac{\rho_{sw}}{\rho_{w}}$

$1.025=\frac{\rho_{sw}}{1000}$

$\rho_{sw}=1025$ kg/m³.

Step2

Difference in pressure is calculated as follows:

$\bigtriangleup p=rho_{sw}gh$

$\bigtriangleup p=1025\times9.81\times320$

$\bigtriangleup p=3217680$ pa.

Or

$\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})$

$\bigtriangleup p=3217.68$ kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

6 0

3 years ago

A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta

Vikki [24]

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore $\theta angle = 0$

and radial component of given velocity is zero

we have $h = r_o v_r_o = 6378+600 =6.97*10^6 m$

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

$\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)$

$GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s$

so

$\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)$

solvingt for $\epsilon)$

$\epsilon = 0.22)$

therefore eccentrcity of orbit is 0.22

6 0

3 years ago