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adelina 88 [10]

2 years ago

6

What do you understand by the statement-The heat of vaporization of water is about 2,260 kJ/kg

2 answers:

Verdich [7]2 years ago

8 0

Answer:The statement "The latent heat of vaporization of water is 226kj/kg." means the amount of heat energy needed to convert one kilogram of water from the liquid phase to vapor is 226kj/kg. Here, the term "latent" refers to a phase-changing process and the term "vaporization" refers to the phase change from liquid to gas

Explanation:

Mice21 [21]2 years ago

3 0

Answer:

The latent heat of vaporization of water is 226kj/kg." means the amount of heat energy needed to convert one kilogram of water from the liquid phase to vapor is 226kj/kg. Here, the term "latent" refers to a phase-changing process and the term "vaporization" refers to the phase change from liquid to gas

Explanation:

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2 years ago

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In a two-body collision, if the momentum of the system is conserved, then which of the

lakkis [162]

Answer:

c) may also be conserved

Explanation:

Momentum is conserved in both elastic and inelastic type of collisions.

But the differences is that:

In an ELASTIC type of collisions, KINETIC ENERGY IS ALSO CONSERVED.

whereas, In an INELASTIC type of collision, KINETIC ENERGY IS NOT CONSERVED.

So unless until type of collision is specified, we can not say anything about the conservation of kinetic energy after collision.

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5 0

3 years ago

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago