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BaLLatris [955]
3 years ago
7

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

Physics
2 answers:
kobusy [5.1K]3 years ago
4 0

Answer: It is the color of the container

Explanation:

The red at the bottom is the color of the bottom of the container. It is not part of the experiment and is not liquid.

The red liquid that is part of the experiment does indeed have the lowest density of the liquids which is why it is floating at the top of al the liquids with the only thing floating on top of it being the blue cube which has a lower density than it.

sdas [7]3 years ago
4 0

Answer: That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.

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AnnZ [28]
What the question for this assessment
8 0
3 years ago
A box has the dimensions of 50 cm × 30 cm × 15 cm, weighs 150 N, and is to be
nadezda [96]

The surface can only withstand a pressure Face A (50cm × 30cm) and Face C (50cm × 15cm) as the surface can only withstand a pressure of 0.25 N/cm³.

<h3>What is pressure?</h3>

The physical force used to apply pressure to an object is defined as such. Per square inch of an object, a force is applied perpendicularly to its surface. For pressure, the fundamental formula is F/A. (Force per unit area). The Pascal is the unit of pressure (Pa).

The four different types of pressure are absolute, atmospheric, differential, and gauge pressure. Have you ever noticed that when you use a straw to drink something, the air actually gets suked out? In reality, you're applying "Pressure" as you drink the beverage.

A box has the dimensions of 50 cm × 30 cm × 15 cm

Let each face be A, B and C

The weight of the box = 150 N

Formula for pressure is

P = F/A

To find out which face of the box can withstand a pressure of 0.25 N/cm

we need find the area of each face and find its pleasure

Face A = 50 cm × 30 cm

Area A = l × b

            = 50 × 30

            = 1500 cm²

Pressure A = 150/1500

                  = 0.1 N/cm³

                    0.25 > 0.1

The surface can definitively withstand the pressure of Face A

Face B = 30 cm × 15 cm

Area A = l × b

            = 30 × 15

            = 450 cm²

Pressure A = 150/450

                  = 0.3 N/cm³

                    0.25 < 0.3

The surface could not withstand the pressure of Face B

Face C = 50 cm × 15 cm

Area A = l × b

            = 50 × 15

            = 750 cm²

Pressure A = 150/750

                  = 0.2 N/cm³

                    0.25 > 0.2

The surface can definitively withstand the pressure of Face C

Thus, The surface can only withstand a pressure Face A (50cm × 30cm) and Face C (50cm × 15cm) as the surface can only withstand a pressure of 0.25 N/cm³.

Learn more about Pressure

brainly.com/question/945436

#SPJ9

8 0
1 year ago
Which activity would be the best choice foir a lifelong fitness program
Fantom [35]

I think jogging/running/walking because you don't need any equipment and you can structure around it on your own time.

7 0
3 years ago
Read 2 more answers
Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig
Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = \frac{1}{3}\pi r^2 h

thus,

volume of the cone = \frac{1}{3}\pi 1.2^2\times 4 = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the \frac{1}{4}times the total height

thus,

center of mass lies at,  h' = \frac{1}{4}\times4=1m

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

4 0
3 years ago
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0
3 years ago
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