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2 years ago

What happens to steam during condensation

Physics

1 answer:

Marta_Voda [28]2 years ago

5 0

Explanation:

Condensation is the process of changing the physical state of matter, from gas to liquid state. During condensation of matter, the physical state of the matter changes. Now in this question it is asked that what happens to the steam during the process of condensation. So steam is the matter, and during this process, steam converts its molecules from gaseous to the liquid state without heating. After this process, the steam is converted back to its liquid state. The liquid now take less space than that of the gas and exert less pressure on the walls of the container.

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An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first

GarryVolchara [31]

Answer:

The right answer is "1.369 m/s²".

Explanation:

The given values are:

Distance (s)

= 260 m

Initial speed (u)

= 26 m/s

Reaction time (t')

= 0.51 s

During reaction time, the distance travelled by locomotive will be:

⇒ $s'=ut'$

$=26\times 0.51$

$=13.26 \ m$

Remained distance between locomotive and car:

⇒ $x=s-s'$

$=260-13.26$

$=246.74 \ m$

Now,

The final velocity to avoid collection is, V = 0 m/s

From third equation of motion:

⇒ $V^2=u^2+2ax$

On putting the estimated values, we get

⇒ $0=(26)^2+2\times a\times 246.74$

⇒ $0=676+493.48a$

⇒ $493.48a=-676$

⇒ $a=-\frac{676}{493.48}$

⇒ $a=1.369 \ m/s^2$

3 0

3 years ago

A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface

givi [52]

Answer:

a = 0.009 J

b = 0.19 m/s

c = 0.005 J and 0.004 J

Explanation:

Given that

Mass of the object, m = 0.5 kg

Spring constant of the spring, k = 20 N/m

Amplitude of the motion, A = 3 cm = 0.03 m

Displacement of the system, x = 2 cm = 0.02 m

Total energy of the system, E =

E = 1/2 * k * A²

E = 1/2 * 20 * 0.03²

E = 10 * 0.0009

E = 0.009 J

E = 1/2 * k * A² = 1/2 * m * v(max)²

1/2 * m * v(max)² = 0.009

1/2 * 0.5 * v(max)² = 0.009

v(max)² = 0.009 * 2/0.5

v(max)² = 0.018 / 0.5

v(max)² = 0.036

v(max) = √0.036

v(max) = 0.19 m/s

V = ±√[(k/m) * (A² - x²)]

V = ±√[(20/0.5) * (0.03² - 0.02²)]

V = ±√(40 * 0.0005)

V = ±√0.02

V = ±0.141 m/s

Kinetic Energy, K = 1/2 * m * v²

K = 1/2 * 0.5 * 0.141²

K = 1/4 * 0.02

K = 0.005 J

Potential Energy, P = 1/2 * k * x²

P = 1/2 * 20 * 0.02²

P = 10 * 0.0004

P = 0.004 J

4 0

2 years ago

ACUTE INFECTION plz help worth 70 pt plz help !!!!! Also brainlest

Paraphin [41]

Answer:

Simple awnser Do it yourself I really would help but I have no clue! Sorry

Explanation:

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3 years ago

Why is it important to understand your strength and weakness<br><br>

S_A_V [24]

Answer: be a better person

Explanation:

6 0

3 years ago

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6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t

spin [16.1K]

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

$p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s$

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

5 0

3 years ago

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