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2 years ago

What is the density of concrete in kilogram per cubic meter.

Physics

2 answers:

Korvikt [17]2 years ago

6 0

Answer:

2400 kg per cubic meter

Explanation:

gulaghasi [49]2 years ago

5 0

Answer:

2400 kilogram per cubic meter

Explanation:

the density of concrete is a measure of its unit weight.

concrete is a mixture of cement, fine and coarse aggregates, water and sometimes some supplementary materials like fly ash, and various admixtures.

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https://www.acs.org/content/acs/en/education/whatischemistry/landmarks/josephpriestleyoxygen.html

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2 years ago

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Why are the Soviets ahead in the race to capture the V-2 rocket and Werner von Braun?

Dahasolnce [82]

The second world war, and its war weapons, such as the v-2 rockets, had a great impact on the world until today, to answer this question we need that...

<h3>V-2 factory </h3>

On April 11, 1945, US troops took the town of Bleicherode, in the Kohnstein region, where the V-2 factory was located. From there about 100 complete V-2s and thousands of parts and equipment were "captured" as war loot and transferred to the United States, where they formed the basis for practical studies of the missile defense program.

With this information, we can say that because it was a base discovered by US troops, none of the alternatives is correct, as it was not the Soviets who discovered it, and that the base was also located in the central part of Germany.

<u>The </u><u>v-2 factory</u><u> was located in </u><u>Kohnstein, central germany,</u><u> by this claim, </u><u>none of the alternatives is correct.</u>

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2 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

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2 years ago

Calculate the gravitational attraction between two objects which masses are: mass A: 2.5kg, mass B: 5kg. The distance between th

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From the calculation, the gravitational force of attraction is 1.33 * 10^-14 N.

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The gravitational force is an attractive force that acts between any two masses.

It is given by;

F = Gm1m2/r^2

F = 6.67 * × 10−11 * 2.5 * 5/(250)^2

F = 83.4 × 10−11 /62500

F= 1.33 * 10^-14 N

Learn more about gravitational force:brainly.com/question/12528243

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Your heart rate is higher than normal.

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