Ask question

Ask question

All categories

2 years ago

8

What happens to particles of a substance as its temperature increases? A the average kinetic energy increases B the average kine

tic energy decreases C the average kinetic energy stays the same D nothing happens

1 answer:

Gemiola [76]2 years ago

6 0

The answer is A. When the the temperature increases the kinetic energy increases

You might be interested in

What is the radius of the Earth?

ki77a [65]

<span>the answer would be 3,959 miles</span>

7 0

2 years ago

2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially

Naddika [18.5K]

Answer:

E_particle = 1,129 10⁻²⁰ J / particle

T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

E_a = 2 5000 = 10000 J

Gas b

E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

E_total = E_a + E_b

E_total = 10000 + 24000 = 34000

The total mass is

M = m_a + m_b

M = 2 +3 = 5

The average energy among the entire mass is

E_averge = E_total / M

E_averago = 34000/5

E_average = 6800 J

One mole of matter has Avogadro's number of atoms 6,022 10²³ particles

Therefore, each particle has an energy of

E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³

E_particle = 1,129 10⁻²⁰ J / particle

For find the temperature let's use equation

E = kT

T = E / k

T = 1,129 10⁻²⁰ / 1,381 10⁻²³

T = 8.175 102 K

T= 817.5 K

5 0

3 years ago

Which sound wave-object interaction is used by animals in echolocation?

tekilochka [14]

The answer is reflection

5 0

3 years ago

Read 2 more answers

1.20 Newton force is working on a 250 gram object. What is the acceleration?

Leokris [45]

Answer:

The answer is B

Explanation:

250g = 0.25kg

F = m × a

a = F/m

= 1.2/0.25

= 4.8m/s²

7 0

3 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

Read 2 more answers