Ask question

Ask question

All categories

2 years ago

14

The closeness of particles of gas and their low speeds allow intermolecular forces to become important at certain pressures and

temperatures. Which best describes this statement?

2 answers:

VladimirAG [237]2 years ago

5 0

Answer:

This is a limitation of the kinetic molecular theory

Explanation:

Lilit [14]2 years ago

4 0

This is a limitation of kinetic-molecular energy. Right?

You might be interested in

Will plane flat mirror will make a real or virtual image?<br> Real<br> Virtual

nikdorinn [45]

Image formed by a plane mirror is always virtual which means that the light rays do not actually come from the image but upright and these of the same shape and size are the object it is<span> reflecting.</span>

8 0

3 years ago

Which of the following is not one of the observed characteristics of our solar system?

Sindrei [870]

<span>The characteristic not observed is the sun and planets rotate in the same direction. The planets in the solar system go around the sun. The sun is in a fixed position relative to planets. The time one planet takes to go around the sun is a year on that planet. the suns gravity keeps the solar system together and th planets revolving aroud it </span>

3 0

3 years ago

Read 2 more answers

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago

Daniel and Christopher were on a plane traveling at a velocity of 200 km/hr south for 1,000 km. How long (in hours) was the plan

docker41 [41]

Answer:
5 hours

step-by-step explanation:
very simple just divide 1,000 and 200 to get 5

3 0

2 years ago

For maximum radiation protection the suggested distance between array fan-beam scanner source and the operator is:

krok68 [10]

For maximum radiation protection the suggested distance between array fan-beam scanner source and the operator is 2m.

The Fan beam 5 position reference system (PRS) uses accurate time-of-flight laser technology to determine vessel position relative to custom reflectors.

A fan beam allows only the measurement of the azimuth angle. A fan beam, one with a narrow beam width in azimuth and a broad beam width in elevation, can be obtained by illuminating an asymmetrical section of the paraboloid.

The operators’ desk should be positioned at least 1m away from a pencil beam, and at least 2m from a fan-beam system. Some older models, that are not now common, require a distance of 3.5 m.

To learn more about scanner here

brainly.com/question/28174696

#SPJ4

4 0

1 year ago