All categories

2 years ago

What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?

Physics

1 answer:

kkurt [141]2 years ago

6 0

Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

$U=\frac{3}{2}nRT$ (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

$U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J$

The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J

You might be interested in

A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to

daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s

Explanation:

It is given that,

Mass of cart 1, $m_1=2\ kg$

Mass of cart 2, $m_2=4\ kg$

Initial speed of cart 1, $u_1=3\ m/s$

Initial speed of cart 2, $u_2=0$ (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{2\times 3}{(2+4)}$

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

$J=F\times t=m(V-u)$

$J=m(V-u)$

$J=2\times (1-3)$

J = -4 kg-m/s

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0

2 years ago

A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiv

zmey [24]

Answer:

(a) With a short line, the A,B,C,D parameters are:

A = 1pu B = 1.685∠60.8°Ω C = 0 S D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 $KV_{LL}$

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

8 0

2 years ago

Which of the following are scalar quantities? select all that apply

dolphi86 [110]

Answer:

1, 4, 5, see the explanation below

Explanation:

We must remember that scalar magnitudes are distinguished by having only a physical quantity, that is, they have no sense or direction as an example of scalar quantities, we find mass, temperature, energy, specific heat, power among others.

1 . 150 [grams] , because is a mass = scalar

4. 5 kilometer [race], is an amount = scalar

5. 34 steps, is an amount = scalar

Number 2, and 3 are vectors because they have amount and direction.

3 0

2 years ago

If the wire is replaced by an infinite current sheet with density Js = 0.40 A/m, what would be the magnetic flux (in T · m2) thr

oksian1 [2.3K]

Answer:

$\phi _{B} =0.855 T-m^{-2}$

Explanation:

given data

density of current sheet = 0.40 A/m

length a = 0.27 m

width b = 0.63 m

For infinite sheet, magnetic field is given as

$B = \mu _{O}J$

magnetic flux is given as

$\phi _{B} = BA$

$= \mu _{O}Jab$

$= 4\pi *0.40*0.27*0.63$

$\phi _{B} =0.855 T-m^{-2}$

6 0

2 years ago

Starting at point 0, you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees. What is your heigh

ira [324]

Answer:

43.58 m

Explanation:

If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees

Using trigonometry ratio

Sin 5 = opposite/hypothenus

Where the hypothenus = 500m

Opposite = height h

Sin 5 = h/500

Cross multiply

500 × sin 5 = h

h = 500 × 0.08715

h = 43.58m

Therefore, the height above the starting point is equal to 43.58m

5 0

2 years ago