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3 years ago

Why does hot air rise?

Physics

2 answers:

Alecsey [184]3 years ago

6 0

It rises because hot air is less dense than cool air.

Dmitriy789 [7]3 years ago

5 0

Difference in the densities.

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6. A balloon inflated with helium is able to float toward the ceiling because

Nostrana [21]

Answer:

b) the density of the balloon is less than the density of the atmosphere.

Explanation:

We must take into account that density is defined as the relationship between mass and volume.

The helium density is 0,1785 [kg/m3]

The air density is 1.2 [kg/m3]

Therefore a substance with lower density (helium) will try to rise on substances with heavier density (air).

An example of this is when we mix water with oil, the water is heavier than the oil, that is, the density of the water is greater than the density of the oil, therefore the oil will be located on the surface of the water and not at the bottom.

Another example is that of a balloon inflated with air is common, this balloon doesn't rise because the air density outside the balloon equals the air density inside the balloon, the balloon will fall to the ground since it contains the weight of the air contains the weight of the material of the balloon.

5 0

3 years ago

How does radiation move?

Degger [83]

Radiation can move through empty space . to transfer heat: convection cannot.

8 0

3 years ago

Calculate the wavelengths of the first five members of the Lyman series of spectral lines, providing the result in units Angstro

Oduvanchick [21]

Answer:

Explanation:

The formula for hydrogen atomic spectrum is as follows

energy of photon due to transition from higher orbit n₂ to n₁

$E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV$

For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc

energy of first line

$E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})$

10.2 eV

wavelength of photon = 12375 / 10.2 = 1213.2 A

energy of 2 nd line

$E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})$

= 12.08 eV

wavelength of photon = 12375 / 12.08 = 1024.4 A

energy of third line

$E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})$

12.75 e V

wavelength of photon = 12375 / 12.75 = 970.6 A

energy of fourth line

$E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})$

= 13.056 eV

wavelength of photon = 12375 / 13.05 = 948.3 A

energy of fifth line

$E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})$

13.22 eV

wavelength of photon = 12375 / 13.22 = 936.1 A

7 0

3 years ago

a 3.0 kg mass moving to the right at 1.4 m/s collides in a perfectly inelastic collision with a 2.0 kg mass initially at rest. w

Sedbober [7]

The velocity of the combined mass after the collision is 0.84 ms-1.

<u>Explanation:</u>

According to law of conservation of momentum, the change in momentum before collision will be equal to the change in momentum of the objects after collision in isolated system.

But as it is perfectly inelastic collision in the present case, the final momentum will be based on the product of total mass of both the object with the velocity with which the collision occurred. This form is attained from the law of conservation of momentum as shown below:

So as law of conservation of momentum,

$M_{1} U_{1}+M_{2} U_{2}=M_{1} V_{1}+M_{2} V_{2}$

Here $M_{1}$ = 3 kg and $M_{2}$ = 2 kg are the masses of objects 1 and 2, $U_{1}$ = 1.4 m/s and $U_{2}$ = 0 are the initial velocities of object 1 and object 2, $V_{1}$ and $V_{2}$ are the final velocities of the objects.

So after collision, object 1 get sticked to object 2 and move together with equal velocity $V_{1}$ = $V_{2}$ = $V_{f}$ . Thus the above equation will become,