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Kruka [31]
3 years ago
15

For a substance to change from a gas to a liquid what must happen to its particles

Physics
2 answers:
baherus [9]3 years ago
5 0

Answer: The answer is A!

Explanation:

Just took the Apex quiz.

mihalych1998 [28]3 years ago
3 0

Answer:

2kg pot at 25C

Explanation:

Apex

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HELP I'll MARK BRAINLY
AURORKA [14]

Answer:

it is plate tectonics option D hope it helps

there are 7 major and 20 minor plates

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3 years ago
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Jack has two boxes: one is 148g and one is 78g. if jack pushes both boxes with the same amount of force which will accelerate fa
kifflom [539]
1) A The 78g
2) C Push on the wagon in the opposite direction as Jack with a force that is the same as Jack is applying.
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3 years ago
An object completes one and half revolution of a circle of radius R calculate the displacement and distance
Papessa [141]

Answer:

Distance is path length covered by particle. When particle moves along half circle, it covers half the circumference therefore distance covered is (2×pi×r)/2 = pi× r. ... Hence displacement is equal to diameter or 2 times the radius of circle.

8 0
3 years ago
A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0
3 years ago
a ball thrown vertically downwatd stikes a horizontal surface with a speed of 15 meters per seconds if then bounces and reaches
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but why was it thrown is the real question

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