Which type of light ray can produce identical but reversed

2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas

Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

$ds=\dfrac{dQ}{dt}$

At constant pressure,

$dQ=C_{p}dt$

$\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}$

$\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})$

$\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}$

$0.693=ln\dfrac{T_{2}}{T_{1}}$

$ln2=ln\dfrac{T_{2}}{T_{1}}$

$T_{2}=2T_{1}$ ...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

$\Delta Q=C_{p}\Delta T$

Put the value into the formula

$6236=2.5\times8.3144(T_{2}-T_{1})$

$6236=20.786(T_{2}-T_{1})$

$T_{2}-T_{1}=\dfrac{6236}{20.786}$

$T_{2}-T_{1}=300$

Put the value of T₂

$2T_{1}-T_{1}=300$

$T_{1}=300\ K$

Put the value of T₁ in equation (I)

$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0

3 years ago

The analysts at Techno InfoSystems are considering the four-model approach to system development for a new client.

garri49 [273]

It will provide a clear picture of current system functions before any modifications or improvements are made is a benefit if they use the four-model approach.

<u>Option: D</u>

<u>Explanation:</u>

The four model approach is followed by number of analyst which showcase that they construct the physical and logical model of both current and new system. The most important advantage of such approach is it portrays transparent image of ongoing system, before one apply any modification or variation. This is necessary because the flaws which generated earlier in system may affect the SDLC phases and outcome of such process may result into unsatisfied user by paying additional cost. This can be avoided by taking additional steps which make it worth it. The major disadvantage of such approach is the added time and cost of constructing model in both current system.

4 0

3 years ago

A farm truck travels due east with a constant speed of 9.50 m/s along a horizontal road. A boy riding in the back of the truck t

Reil [10]

Answer:

a)he angle is 90 with respect to the horizontal (x axis)

b) its speed is zero both vertically and horizontally

c) vertical path

d) a parabolic movement

e) v₀ = (9.5i ^ + 8.232 j ^) m / s

Explanation:

This is a relative problem and movement.

.v ’= v + u

Where v ’is the speed with respect to the mobile system, v the speed with respect to the fixed system and the speed between the two reference systems

a) The child and the can is in the truck, so they go at the speed of the truck, when he throws the can he continues at this speed on the x-axis and therefore as the two advance the same distance the more hands of the child, consequently the can is thrown vertically

The angle is 90 with respect to the horizontal (x axis)

b) with respect to the truck, the can is still, so its speed is zero both vertically and horizontally

c) The child sees that the can follows a vertical path

d) A stationary observer on the ground, sees that the can has a constant speed in the same direction of the truck and when they throw it vertical goal has a vertical movement, the sum of these two movements gives a parabolic movement of the same uncle as a projectile launch

e) the initial speed has two components

X Axis v_lata = v_camion = 9.5 m / s

Y Axis speed given by the child

Let's look for the travel time

x = v₀ₓ t

t = x / v₀ₓ

t = 16 / 9.5

t = 1.68 s

y = $v_{oy}$ t - ½ g t²

When he returns to the child's hand and = 0

0 = v_{oy} t - ½ g t²

v_{oy} = ½ g t

v_{oy} = ½ 9.8 1.68

v_{oy} = 8,232 m / s

Speed is

v₀ = (9.5i ^ + 8.232 j ^) m / s

5 0

3 years ago

Helium atoms have a mass of 4u and oxygen molecules have a mass of 32u, where u is defined as an atomic mass unit (u=1.660540 x

Elden [556K]

Answer:

300 K

37.5 K

Explanation:

$k_b$ = Boltzmann constant = $1.38\times 10^{-23}\ J/K$

T = Absolute temperature

$M_h$ = Mass of Helium atom = 4 u

$M_o$ = Mass of Oxygen atom = 32 u

$u=1.660540\times 10^{−27}\ kg$

The kinetic energy of a gas is given by

$K=\frac{3}{2}k_bT$

So, the temperature would be 300 K

RMS velocity is given by

$v=\sqrt{\frac{3k_bT}{M}}$

$\frac{T_h}{M_h}=\frac{T_o}{M_o}\\\Rightarrow T_h=\frac{T_o}{M_o}\times M_h\\\Rightarrow T_h=\frac{300\times 4}{32}\\\Rightarrow T_h=37.5\ K$

The gas temperature is 37.5 K

5 0

3 years ago

A factory worker pushes a 30.0kg crate a distance of 4.4 m along a level floor at constant velocity by pushing downward at an an

andre [41]

Answer:

1) F = 94.58 N , 2) W1 = 360.4 J , 3) W2 = -360.4 J , 4) W3 =0

Explanation:

1- To find the value of the force, let's use Newton's second law, where in the x-axis we have the applied force and the friction force that opposes the movement and the axis and we have the normal and the weight, look at the attached to see A free body diagram.

We see that the applied force (F) must be decomposed

Cos θ = Fx / F

Fx = f cos θ

sinθ = Fy / F

Fy = F sin θ

As the box moves at constant speed the acceleration is zero

X axis

Fx-fr 0 = 0

Fx = fr

fr = μ N

F cos T = μ N

Axis y

N -W- Fy = 0

N- F sin θ = mg

Let's write the equation system and solve it

F cos θ = μ N

N - F sin θ = mg

N = mg + F sinθ (1)

F cos θ = μ (mg + F sin θ)

F (cos θ - μ Sin θ) = μ mg

F (cos 30 - 0.24 sin 30) = 0.24 30.0 9.8

F = 70.56 / 0.746

F = 94.58 N

This is the value of the force applied

2- The work is defined as the scalar product of the force by the distance traveled, in this case as we have the components of the force the only one that performs work is the Enel X-axis component, because with the other the angle is 90º and the cosine of 90 is zero

W1 = Fx X

W1 = 94.58 cos 30 4.4

W1 = 360.4 J

3- Let's calculate the value of the friction force

fr = μ N

The value of normal can be found in equation 1

N = mg + F sin θ

N = 30.0 9.8 + 94.58 sin 30

N = 341.3 N

fr = 0.24 341.3

fr = 81.9 N

Now we can calculate the work, but let's look at the angle between the displacement and the friction force that is 180º so the cosine of 180 = -1

W2 = - fr X

W2 = - 81.9 4.4

W2 = -360.4 J

4- The normal one has a direction perpendicular to the displacement, so its angle is 90º and the cos 90 = 0, which implies that the work is zero

W3 =0

5- To calculate total work we just have to add the work of each force

W all = W1 + W2 + WN + Ww

W all = 360.4 + (- 360.4) + 0+ 0

W all = 0

4 0

3 years ago