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zhuklara [117]
4 years ago
5

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

meters long, 80 meters wide, and 40 meters high, determine how far the bottom of the ship is below sea level. 4. [psea 1 020 kgm3 =
a) 10 m
b) 15 m
c) 20 m
d) 25 m
e) 30 m
Physics
1 answer:
IrinaVladis [17]4 years ago
7 0

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.

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Whereas in an endothermic reaction, heat is supplied from outside and absorbed by the reactant molecules. Hence, enthalpy of the system increases.

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The cricket player while catches the ball wears gloves and why​
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3 years ago
Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of
mylen [45]

Answer:

density = 5520 kg/m^3

Explanation:

given that

radius of earth = 6378 km

G = 6.67 x 10⁻¹¹ m³/kg.s²

g = 9.80 m/s²

we know,

g = \dfrac{GM}{r^2}

mass of earth

M = \dfrac{gr^2}{G}

M = \dfrac{9.8 \times (6378 \times 10^3)^2}{6.67 \times 10^{-11}}

M = 5.972 x 10²⁴ kg

density =\dfrac{mass}{volume}

V = volume of the earth = 4/3πr³

V = 4/3 x 3.14 x (6378  x 10³)³

V = 1.08 x 10²¹ m³

density = \dfrac{5.972\times 10^{24}}{1.08\times 10^{21}}

density = 5.52 x 10³  kg/m^3

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3 years ago
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

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Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
3 years ago
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