Which process can separate out the solute from the solvent?

A bus covers 10 km in 7 minutes. Find the speed of the bus in km/h

Shtirlitz [24]

Answer:

Explanation: so how many minutes are in an hour 60 right, and the bus travels 10km in 7 minutes right so use math the bus travels 14km in 10 minutes so the bus travels 98km in an hour

8 0

3 years ago

A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0

3 years ago

Distance versus Displacement Worksheet

Umnica [9.8K]

when we find the distance we will add all the blocks so

distance = 6+6+4

distance = 14blocks

when we find the displacement we will add and minus too

As you can read he goes to the south 6 and to north 6 so he leave that place and back to the place again so the displacement is 0. and again he goes to the west 4 blocks so the displacement = 4blocks to the west

6 0

4 years ago

Read 2 more answers

A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa

pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

M= I A

M= $16 \times 10^{-3} \times \pi \times r^2$

M= $16 \times 10^{-3} \times \pi \times 0.3024^2$

M= $4.59 \times 10^{-3}\ A m^2$

b) torque exerted in the loop

$\tau = M\ B$

$\tau = 4.59 \times 10^{-3}\times 0.79$

$\tau = 3.63 \times 10^{-3} N.m$

8 0

3 years ago

The six statements below represent Newton's three laws of motion and Kepler's three laws of planetary motion. Match each stateme

mote1985 [20]

Answer:

1. Force = mass x acceleration - Newton

2. A planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out equal areas in equal times - Kepler

3. For any force, there is an equal and opposite reaction force - Newton .

4. An object moves at constant velocity if there is no net force acting upon it - Newton

5. The orbit of each planet about the Sun is an ellipse with the Sun at one focus - Kepler.

6. More distant planets orbit the Sun at slower average speeds, obeying the precise mathematical relationship p2-a3 - Kepler.

Explanation:

The three laws of planetary motion formulated by Johannes Kepler or Kepler's laws of planetary motion:

The first law states that the planets move around the Sun in an elliptical orbit with the Sun at one of the foci.
The second law states that the line segment joining a planet to the Sun sweeps out equal areas in equal time.
The third law states that the square of the orbital period (p) of a planet is directly proportional to the cube of the mean distance (a) from the Sun (or semi-major axis of its orbit) i.e., p² is proportional to a³.

The three laws of motion formulated by Sir Isaac Newton or Newton's laws of motion:

The first law, also known as the law of inertia states that an object at rest or moves at a constant velocity will remain at rest or keep moving at a constant velocity unless it is acted upon by a force.
The second law states that the total force (F) applied on an object is directly related to the acceleration (a) of that object produced by the applied force and the mass (m) of the object, i.e., F = ma (assuming the mass m is constant).
The third law, also known as the law of action and reaction states that when an object exerts a force on another object, then the latter exerts a force equal in magnitude and opposite in direction on the former object i.e., for every action, there is an equal and opposite reaction. The example includes the recoiling of a gun when it fires a bullet forward.

5 0

4 years ago