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3 years ago

Suppose a log's mass is 5 kg. After burning, the mass of the ash is 1 kg. explain what May have happened to the other 4 kg

1 answer:

4 0

I think the "missing mass" may have escaped the site of the
conflagration in the form of hot, invisible gases, and in the form
of all the tiny particles in the smoke.

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What happens when an ionic compound is melted?

Dahasolnce [82]

It conducts electricity. This is because the electrons are able to move around more for it has gained energy, after being heated.

7 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

A ferry coming into port is sailing at 12 m/s. It takes 2.5km to come to rest in the port. Calculate the deceleration of the fer

pogonyaev

Answer: -0.0288 m/s^2

Explanation:

Let's suppose that the ferry decelerates at a constant rate A (deceleration is an acceleration in the opposite direction to the original motion of an object)

Then the acceleration equation of the ferry will be:

a(t) = -A

(the negative sign is because this acceleration is in the opposite direction with respect to the movement of the ferry)

To get the velocity equation of the ferry, we need to integrate with respect to the time, t, we will get:

v(t) = -A*t + v0

where v0 is the initial velocity of the ferry, v0 = 12m/s.

v(t) = -A*t + 12m/s

For the position equation of the ferry we need to integrate again over time:

p(t) = (-A/2)*t^2 + (12m/s)*t + p0

Where p0 is the initial position of the ferry, in this case, it can be zero, because it will depend on where we put the origin on our coordinate axis.

then p0 = 0m

P(t) = (-A/2)*t^2 + (12m/s)*t

The ferry will come to rest at the moment when it's velocity is equal to zero, this will happen when:

v(t) = 0m/s = -A*t + 12m/s

We need to find the value of t.

A*t = 12m/s

t = (12m/s)/A

Now we can replace this in the position equation because we know that the ferry needs 2.5 km or 2500 meters to come to rest.

p( (12m/s)/A) = 2500m = (-A/2)*( (12m/s)/A)^2 + (12m/s)*((12m/s)/A)

2500m = (-72 m^2/s^2)/A + (144m^2/s^2)/A

2500m = (72 m^2/s^2)/A

2500m*A = (72 m^2/s^2)

A = (72 m^2/s^2)/2500m = 0.0288 m/s^2

and the acceleration of the ferry was -A, then the acceleration of the ferry is:

-0.0288 m/s^2

4 0

3 years ago

Why does the current is reduced as electrons move through a conductor

hammer [34]

Because the electrons collide with the particles inside the conductor so are therefore slowed down seen as current is the rate of flow of electrons

3 0

3 years ago

A student drops a ball from the top of a 10-meter tall building. The ball leaves the thrower's hand with a zero speed. What is t

Sergio [31]

Answer:

14 m/s

Explanation:

u = 0, h = 10 m, g = 9.8 m/s^2

Use third equation of motion

v^2 = u^2 + 2 g h

Here, v be the velocity of ball as it just strikes with the ground

v^2 = 0 + 2 x 9.8 x 10

v^2 = 196

v = 14 m/s

7 0

3 years ago