Which is described as the flow of charge particles?

I a liter jar contains 0.5 moles of gas at a pressure of 2 ATM what is the temperature of the gas

sweet [91]

Answer:

<h2>The temperature of the gas is 48.75 Kelvin.</h2>

Explanation:

Using the ideal gas equation as shown

PV = nRT where;

P is the pressure of the gas in ATM

V is the volume of the gas

n is the number of moles

R is the ideal gas constant

T is the temperature in Kelvin

From the formula, $T = \frac{PV}{nR}$

Given the following parameters V = 1litre, n = 0.5moles. pressure = 2ATM

R = 0.08206 atm L/molK

On substituting to get the temperature we have:

$T = \frac{2*1}{0.08206*0.5} \\T = \frac{2}{0.04103}\\T = 48.75Kelvin$

5 0

3 years ago

A skateboarder shoots off a ramp with a velocity of 7.3 m/s, directed at an angle of 60° above the horizontal. The end of the ra

erica [24]

Answer:

y maximum 3.54 m, value X 2.35 m

Explanation:

We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero

Vyf² = Vyo² - 2g (Y-Yo)

Where Yo is the initial height of the ramp 1.5 m

0 = Vyo² -2g (Y-Yo)

Y-Yo = Voy² / 2g

Y = Yo + Voy² / 2g

Let's calculate the velocity components using trigonometry

Voy = vo without T

Vox = Vo cost

Voy = 7.3 sin 60

Vox = 7.3 cos 60

Voy = 6.32 m / s

Vox = 3.65 m / s

Let's calculate the maximum height

Y = 1.5 +6.32²/2 9.8

Y = 3.54 m

This is the maximum height from the ground

b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point

Vfy = Voy - gt

0 = Voy - gt

t = Voy / g

t = 6.32 / 9.8

t = 0.645 s

Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time

X = Vox t

X = 3.65 0.645

X= 2.35 m

8 0

3 years ago

An object in equilibrium has a net force of

Butoxors [25]

Answer:

An object in equilibrium has a net force of zero

Static equilibrium describes an object at rest having equal and balanced forces acting upon it.

Dynamic equilibrium describes an object in motion having equal and balanced forces acting upon it.

Explanation:

An object is said to be in equilibrium when a net force of zero is acting on it. When this condition occurs, the object will have zero acceleration, according to Newton's second law:

$F=ma$

where F is the net force, m the mass of the object, a the acceleration. Since F=0, then a=0. As a result, we have two possible situations:

- If the object was at rest, then it will keep its state of rest. In this case, we talk about static equilibrium.

- If the object was moving, it will keep moving with constant velocity. In this case, we talk about dynamic equilibrium.

8 0

3 years ago

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

katen-ka-za [31]

Answer:b

Explanation:

Given

mass of first cart $m_1=6 kg$

mass of second cart $m_2=3 kg$

velocity of first cart $v_1=3 m/s$

conserving momentum

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

$6\times 3+3\times 0=(9)\cdot v$

$v=2 m/s$

Initial kinetic Energy $K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$

$K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0$

$K.E._1=27 J$

Final Kinetic Energy

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J$

Ratio of initial Kinetic Energy to the Final Kinetic Energy

$=\frac{27}{18}=1.5$

6 0

4 years ago

The rate of spin of a disk is decreasing. A point on the disk that is near the rim of the disk will have...

NNADVOKAT [17]

At a point near the rim of the disk, it will have a<span> non-zero radial acceleration and a zero tangential acceleration. Also known as centripetal acceleration, radial acceleration takes place along the radius of the disk. On the other hand, the tangential acceleration is along the path of disk's motion.</span>

8 0

3 years ago