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4 years ago

What do we call changes between solid liquid and gaseous forms of a substance

Physics

2 answers:

Nadusha1986 [10]4 years ago

6 0

Answer:

a change of state

Explanation:

const2013 [10]4 years ago

5 0

We call changes between solid liquid and gaseous forms of a substance as phase change or change of state.

Explanation:

To change a substance from one state to another, extreme temperatures or pressures are required. Sometimes when a substance doesn't change states we should use all the ideas when that happens. To create a solid, we should decrease the temperature by a huge amount and then add pressure. For example, oxygen will solidify at -361.8 degrees Fahrenheit at standard pressure. However, it will freeze at warmer temperatures when the pressure is increased.

Phase changes happen when a substance reach some special points. Sometimes when a liquid becomes a solid a freezing point or melting point is used to measure the temperature at which a liquid changes into a solid. Some of the phase changes are: Condensation, Freezing, Melting.

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Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object

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The area of the rectangle plus the area of the triangle under the line .-.

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4 years ago

Read 2 more answers

Three samples of gas each exert 740. mm Hg in separate 2 L containers. What pressure do they exert if they are all placed in a s

Usimov [2.4K]

According to Dalton's law of partial pressure, the total pressure exerted is simply equal to the sum of the partial pressures of the individual gases. Given that all three samples of gas each exert 740 mmHg, when they are placed in a single 2 L container, they exert a pressure of 2220 mmHg on the container which is the sum of their individual pressures.

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3 years ago

School homework about multiplying fractions worth 30 brainly points

RoseWind [281]

Answer:

Explanation:

1. $\frac{1}{3}$ x $\frac{6}{7}$ = $\frac{6}{21}$

= $\frac{2}{7}$

b. $\frac{6}{8}$ + $\frac{4}{9}$ = $\frac{24}{72}$

= $\frac{1}{3}$

c. $\frac{10}{15}$ x $\frac{3}{4}$ = $\frac{30}{60}$

= $\frac{1}{2}$

d. $\frac{7}{10}$ of $\frac{5}{10}$ = $\frac{7}{10}$ x $\frac{5}{10}$

= $\frac{35}{100}$

= $\frac{7}{20}$

e. $\frac{3}{8}$ of $\frac{4}{6}$ = $\frac{3}{8}$ x $\frac{4}{6}$

= $\frac{12}{48}$

= $\frac{1}{4}$

f. $\frac{7}{12}$ of $\frac{9}{14}$ = $\frac{7}{12}$ x $\frac{9}{14}$

= $\frac{63}{168}$

= $\frac{3}{8}$

2. $\frac{22}{6}$ x $\frac{3}{11}$ = $\frac{66}{66}$

= 1

b. $\frac{15}{6}$ x $\frac{4}{5}$ = $\frac{60}{30}$

= 2

c. $\frac{25}{8}$ x $\frac{4}{10}$ = $\frac{100}{80}$

= $\frac{5}{4}$

d. $\frac{33}{12}$ x $\frac{4}{15}$ = $\frac{132}{180}$

= $\frac{11}{15}$

4 0

3 years ago

A car with 2 × 10^3 kg moving at a speed of 10 m/s collides and sticks with car B of mass of 3 × 10^3 kg initially at rest. How

stepan [7]

Answer:

$6 \times 10^4 \; \rm J$ .

Explanation:

KE lost = Total KE before Collision - Total KE after Collision.

For each car, the KE before collision can simply be found with the equation:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ , where

$m$ is the mass of the car, and
$v$ is the speed of the car.

The $2 \times 10^3\; \rm kg$ car would have an initial KE of:

$\displaystyle \frac{1}{2} \times 2 \times 10^3 \times 10^2 = 10^5\; \rm J$ .

The $3 \times 10^3\; \rm kg$ car was initially not moving. Hence, its speed and kinetic energy would zero before the collision.

To find the velocity of the two cars after the collision, apply the conservation of momentum.

The momentum $p$ of an object is equal to its mass $m$ times its velocity $v$ . In other words, $p = m\cdot v$ .

Let the mass of the two cars be denoted as $m_1$ and $m_2$ , and their initial speeds $v_1$ and $v_2$ . Since the two cars are stuck to each other after the collision, their final speeds would be the same. Let that speed be denotes as $v_3$ .

Initial momentum of the two-car system:

$\begin{aligned}& m_1 \cdot v_1 + m_2 \cdot v_2 \\ &= 2 \times 10^3 \times 10 + 3 \times 10^3 \times 0 \\ &= 2 \times 10^4\; \rm kg \cdot m \cdot s^{-1}\end{aligned}$ .

After the collision, both car would have a velocity of $v_3$ (since they were stuck to each other.) As a result, the final momentum of the two-car system would be:

$m_1\cdot v_3 + m_2 \cdot v_3 = (m_1 + m_2)\, v_3$ .

Since momentum is conserved during the collision, the momentum of the system after the collision would also be $2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ . That is: $(m_1 + m_2) \, v_3 = 2 \times 10^4 \; \rm kg \cdot m \cdot s^{-1}$ .

Solve for $v_3$ :

$\begin{aligned} v_3 &= \frac{(m_1 + m_2)\, v_3}{m_1 + m_2} \\ &= \frac{2 \times 10^4}{2 \times 10^3 + 3 \times 10^3} \\ &= \frac{2 \times 10^4}{5 \times 10^3} \\ &= 4 \; \rm m \cdot s^{-1}\end{aligned}$ .

Hence, the total kinetic energy after the collision would be:

$\begin{aligned} &\frac{1}{2}\, m_1 \, v^2 + \frac{1}{2}\, m_2\, v^2 \\ &= \frac{1}{2}\, (m_1 + m_2)\, v^2 \\ &= \frac{1}{2} \times \left(2 \times 10^3 + 3 \times 10^3\right) \times 4^2 \\ &= 4 \times 10^4\; \rm J\end{aligned}$ .

The amount of kinetic energy lost during the collision would be:

$\begin{aligned}&\text{KE After Collision} - \text{KE Before Collision} \\ &= 10^5 - 4 \times 10^4 \\&= 6\times 10^4\; \rm J \end{aligned}$ .

5 0

3 years ago

A girl standing upright exerts a pressure of 15000 N/m2 on the floor. Given that the total area of contact of shoes and the floo

lapo4ka [179]

Explanation:

F=15000/0.02

=300N

total area contact of the shoes and floor when stood in one foot=0.02m^2/2=0.01m^2

P=F/A

=300/0.01

=30 000Pa/30 000Nm^-2

in another way=15000×2

=30000N/m^2

5 0

4 years ago