True or False: The variables in the equation 4x-(5y)2=64x-(5y)2=6 are 4, 5, and 6

A supertanker filled with oil has a total mass of 6.1 x108 kg. If the dimensions of the ship are those of a rectangular box 300

IrinaVladis [17]

Answer:

The bottom of the sea is 25 m below sea level.

Explanation:

Given data

Mass = 6.1 × $10^{8} \ kg$

$\rho_{sea} = 1020\ \frac{kg}{m^{3} }$

We know that Buoyant force on the tank is equal to gravity force of the tank.

$F_B = F_g$

$(\rho_{Fluid}) (g) (V_{disp}) = m g$

$(\rho_{Fluid}) (V_{disp}) = m$

1020 × $V_{disp}$ = 6.1 × $10^{8}$

$V_{disp}$ = 598039.21 $m^{3}$

We know that

$V_{disp}$ = W × L × H

598039.21 = 300 × 80 × H

H = 25 m

Therefore the bottom of the sea is 25 m below sea level.

7 0

3 years ago

A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w

balu736 [363]

Wow ! This one could have some twists and turns in it.
Fasten your seat belt. It's going to be a boompy ride.

-- The buoyant force is precisely the missing 30N .

-- In order to calculate the density of the frewium sample, we need to know
its mass and its volume. Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

 Weight = (mass) x (gravity)
 185N = (mass) x (9.81 m/s²)
 Mass = (185N) / (9.81 m/s²) = 18.858 kilograms of frewium

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³. So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

 30N = (mass of the displaced water) x (9.81 m/s²)

 Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

 Volume of displaced water = 3,058 cm³

Finally, density of the frewium sample = (mass)/(volume)

 Density = (18,858 grams) / (3,058 cm³) = 6.167 gm/cm³ (rounded)

================================================

I'm thinking that this must be the hard way to do it,
because I noticed that

 (weight in air) / (buoyant force) = 185N / 30N = 6.1666...

So apparently . . .

 (density of a sample) / (density of water) =

 (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.

3 0

3 years ago

A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c

tatuchka [14]

We have that the values for F north, F east, F up are