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vovikov84 [41]
4 years ago
13

The block in the illustration is floating in water. Water has a density of 1.0 g/cm 3 and about 25 percent of the block is below

water. What is a good estimate of the density of the block?
Physics
2 answers:
expeople1 [14]4 years ago
7 0
0.75 g/cm3 

let me know if I'm right~! =^-^=  hope it helps~!

~Cecildoesscience
jonny [76]4 years ago
3 0

Answer;

0.25 g/cm^3

Explanation;

25 % of 1 g/cm^3

=  0.25 g/cm^3

Water's density is 1 g/ml. Objects with a density greater than 1g/ml will sink when placed in water.

Objects with a density less than one will float when placed in water. If a block has a density of 0.1g/cm3 it will float because it's density is less than 1.

The percent of the block that will be below the water depends on the density of the floating object. To find the percentage of the object below the water we just convert the density of the object as a percentage.

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1. A student gathered two boxes of the same size made of different materials: glass and clear plastic. She placed them on a wind
Kobotan [32]

Answer:

1.  to relate the type of box material to the temperature of the air within the box

2. Question is incomplete

3. scatterplot

Explanation:

1. The only thing done differently in this experiment is the type of material used in making the boxes, hence the experiment must be about that. Before proceeding to answering this question, we must have this at the back of our minds.

We can gather from the experiment that the boxes are of the same size and were subjected to sunlight for an hour (the same time duration for both). Hence, the temperature of the air inside the box will only be affected by the type of material the box is made of since the boxes have the same size and were subjected to sunlight for the same duration.

From the options provided, the best description for this experiment is; to relate the type of box material to the temperature of the air within the box.

2. The question is incomplete. The value for speed/velocity needed to calculate the average time is missing.

However, the formula needed here is velocity = distance ÷ time

3. There are two variables in this experiment; distance and time

The type of graph that shows two variables on it (of the options provided) is the scatterplot.

8 0
4 years ago
Three liquids are at temperatures of 6 ◦C, 23◦C, and 38◦C, respectively. Equal masses of the first two liquids are mixed, and th
kupik [55]

The equilibrium temperature is T13=3.12 ◦C

<u>Explanation:</u>

<u>Given </u>

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12= 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13= ??

<u>1.When the first two liquids are mixed:</u>

  • mC1(T1-T12)+mC2(T2-T12)=0
  • C1(6-13)=C2(23-13)=0
  • 7C1=10C2
  • C1=1.42C2

<u>2.When the second and third liquids are mixed</u><u>:</u>

  • mC2(T2-T23)+mC3(T3-T23)=0
  • C2(23-26.8)=C3(38-26.8)=0
  • 3.8C2=12.8C3
  • C2=3.36C3

<u>3.When the first and third liquids are mixed:</u>

  • mC1(T1-T13)+mC3(T3-T13)=0
  • C1(6-T13)+C3(38-T13)=0
  • C1=1.42C2  C2=3.36C3
  • C1=1.42C2(3.36C3)
  • C1=4.77C3
  • C1(6-T13)+C3(38-T13)=0
  • 4.77C3(6-T13)+C3(38-T13)=0
  • By solving the equation we get,
  • T13=3.12 ◦C
  • The equilibrium temperature is T13=3.12 ◦C

<u></u>

7 0
3 years ago
PLEASE HELPPP MEEE :((​
Hoochie [10]
Power = work / time --> time = work / power = 3600 J / 275 watts = 13.1 seconds.

Make me brainliest plz
4 0
3 years ago
A third class lever has a mechanical advantage of &lt;1. What is an example of a third class lever and why use it?
timurjin [86]

Answer: Choice B: baseball bat; increases velocity

Explanation:

I got this right on my test !!!!!!!

6 0
4 years ago
Read 2 more answers
f 720-nm and 620-nm light passes through two slits 0.68 mm apart, how far apart are the second-order fringes for these two wavel
valkas [14]

Answer:

0.0003 m = 0.3 mm

Explanation:

For constructive interference in the Young's experiment.

The position of the mth fringe from the central fringe is given by

y = L(mλ/d)

λ = wavelength = 720 nm = 720 × 10⁻⁹ m

L = distance between slits and screen respectively = 1.0 m

d = separation of slits = 0.68 mm = 0.68 × 10⁻³ m

m = 2

y = 1(2 × 720 × 10⁻⁹/(0.68 × 10⁻³) = 0.00212 m = 2.12 mm

For the 620 nm light,

y = 1(2 × 620 × 10⁻⁹/(0.68 × 10⁻³) = 0.00182 m = 1.82 mm

Distance apart = 2.12 - 1.82 = 0.3 mm = 0.0003 m

8 0
3 years ago
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