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pickupchik [31]
3 years ago
8

Changing states of matter

Physics
1 answer:
guajiro [1.7K]3 years ago
5 0

Answer:

....,..,.....,....................solid liquid gas

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What is the time period of a simple harmonic oscillator with a mass of 0.3kg and a force constant of 5N/m?
ikadub [295]

Answer: 6N/m

Explanation:

8 0
4 years ago
4. An object is moving with an initial velocity of 9 m/s. It accelerates at a rate of 1.5
Mazyrski [523]

Answer:

11.87 ms⁻¹

Explanation:

You can use the kinematic equation

v² = u² + 2as

Where v = final velocity

          u = initial velocity

          a = acceleration

          s = displacement

v² = 9² + 2×1.5×20

So you get v = 11.87 ms⁻¹

7 0
3 years ago
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v_{1} be the velocity of ball A along positive x axis

v_{2} be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v_{1}

∴  v_{1} = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v_{2}

2 = u +  v_{2} → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 =  v_{2} - u → equation 2

By adding both equations 1 and 2 we get

v_{2} = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
4 years ago
How we can best share the observation of the light through one hole the paper
zhenek [66]
Drawing is an easy way to connect to your inner child

5 0
3 years ago
One object has a charge of +5.0 · 10-6 c, and a second object has a charge of +2.0 · 10-6
ruslelena [56]
The electrostatic force between two charged objects is given by
F=k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 is the charge of the first object
q2 is the charge of the second object
r is the separation between the two objects

In our problem:
q_1=+5.0 \cdot 10^{-6} C
q_2 = +2.0 \cdot 10^{-6} C
r=0.5 m
So if we plug these numbers into the equation, we can find the electrostatic force between the two objects:
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(5.0 \cdot 10^{-6} C)(2.0 \cdot 10^{-6}C)}{(0.5 m)^2}=0.36 N
4 0
3 years ago
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