PLEASE HELP ANSWER THIS QUESTION !

vfiekz [6]

The density of the fluid is 776.3 $m^{-3}$

Explanation:

Buoyant force is the upward pushing force whenever an object is trying to get immersed in fluid. So this is the force given by the fluid on the object which is trying to get immersed. The buoyant force is found to be directly proportional to the product of density of the object, volume of the object. And here the acceleration due to gravity will be acting as proportionality constant.

$Buoyant force = Density \times Volume \times Acceleration$

As, buoyant force is given as 671 N and volume is 0.0882 $m^{3}$ and acceleration is known as 9.8 m/ $s^{2}$ . Then density is

$\text { Density }=\frac{\text { Buoyant force }}{\text {Volume } \times \text {Acceleration}}$

Thus,

$\text { Density }=\frac{671}{0.0882 \times 9.8}=\frac{671}{0.86436}=776.296 \mathrm{kg} / \mathrm{m}^{3}$

Density is 776.3 kg $m^{-3}$ .

7 0

3 years ago

Alkaline earth metals have a low density true false

Marta_Voda [28]

true

Explanation:

this is because melting point and boiling point decreases down the group because they are held together by attractions between positive nuclei and delocalised electrons

6 0

3 years ago

During a supernova, the outer layers of a star are blown off and the star's core shrinks down by a factor of 10,000 or more to f

Lana71 [14]

Answer:

a. the core will spin faster.

Explanation:

By law of conservation of angular momentum

(mvR)i= (mvR)f

m= mass of star

v= speed of star

R= radius of star

i= initial

f= final

since, size(R) of the star is reduced by factor of 10,000 and mass remains the same, the velocity must increase by the same factor to keep the angular momentum conserved.

Hence, a. the core will spin faster.

8 0

4 years ago

slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a

Ostrovityanka [42]

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth d and the other (lets call it l) is given by:

$l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft$

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section A at that point:

$A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}$

And the raising speed v of the water is given by:

$v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}$

where q is the water flow (1 cubic foot per minute).

7 0

3 years ago

The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur

Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

P₂ is the pressure in the hose

ρ is the density of the fluid

h₁ is the initial height

h₂ is the final height

v₁ is the initial velocity of the fluid

v₂ is the final velocity of the fluid and

g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

= (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

= (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

= (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

= 5.92 × 10⁵ Pa

= 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

brainly.com/question/9506577

#SPJ4

6 0

2 years ago