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andrezito [222]
2 years ago
8

True or false Water is a solute in moist air

Physics
2 answers:
Advocard [28]2 years ago
8 0

Answer:

True

Explanation:

Oxana [17]2 years ago
3 0

Answer:

its true water is a solute in moist air

Explanation:

i hope i helped

a crown is always appreciated

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Describe a real-world example of the law of conservation of momentum.
mario62 [17]
Imagine a car crash. A car coming at a high speed has a head on collision with a car at rest. When the car makes impact, it will move the other car with it at a slower speed then it was travelling at. In this case, the velocity decreased since the car slowed down, but the mass increased since there are now two cars moving. Momentum was conserved because the change in mass accounts for the loss of velocity.
4 0
2 years ago
If a 20 N block experiences a force of kinetic friction of 8.0 N
shutvik [7]

The coefficient of kinetic friction (μ) between the block and the table is 0.4.

<h3>What is kinetic friction?</h3>

This sis the frictional force between an object in motion with the surface in contact.

μN = ff

where;

  • N is normal reaction due to weight of the block
  • ff is frictional force
  • μ is coefficient of friction

μ = ff/N

μ = 8/20

μ = 0.4

Thus, the coefficient of kinetic friction (μ) between the block and the table is 0.4.

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ1

6 0
1 year ago
A stone tumbles into a mine shaft and strikes bottom after falling for 4.2 second how deep is the mine shaft
rjkz [21]
4.2*9.8\\41.16

41.16 meters
5 0
3 years ago
Caculate the total charge transferred through a wire carrying a current of 0.5 Ameter for a period of 20 seconds.
Marina CMI [18]

charge = current x time = 0.5x 20=10Coulombs

5 0
3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
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