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3 years ago

The procedure that scientists should follow when investigating nature

1 answer:

3 0

1. Define a question to investigate
2. Make a hypothesis (prediction)
3. Gather data from outside
4. Analyse the data
5. Draw a conclusion and see if it fits with the hypothesis
6. See how you can improve the experiment

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Benzene gas (C6H6) at 25° C and 1 atm, enters a combustion chamber operating at steady state and burns with 95% theoretical air

tiny-mole [99]

Answer:

Explanation:

Benzene gas is burned with 95 percentage theoretical air during a steady â flow combustion process. The mole fraction of the CO in the products and the heat transfer from the combustion chamber are to be determined

Assumption

steady operating conditions exit .

Air and combustion gases are gases

Kinetic and potential energies are negligible

The fuel is burned with insufficient amount of air and thus the products will contain some CO as well as CO2, H2O and H2

Combustion equation is

C6H6 + ath (O2 + 3.76N2) â 6CO2 +3H2O +3.76ath N2

Where ath is the stoichiometric coefficient and is determined from the O2 balance

ath = 6+1.5 = 7.5

then the actual combustion equation can be written as

C6H6 + 0.95 * 7.5(O2 + 3.76N2) â xCO2 + (6-x) CO + 3H2O + 26.79N2

O2 balance: 0.95 * 7.5 = x +(6-x)/2 +1.5 â x=5.25

Thus C6H6 + 7.125(O2 +3.76 N2) â 5.25 CO2 + 0.75CO + 3H2O + 26.79N2

The mole fraction of CO in the products is

yCO = N CO/ N total = 0.75/5.25 + 0.75+3+26.79

=0.021 or 2.1% mole fraction of the CO in the products

5 0

4 years ago

Read 2 more answers

What is the air change rate (ACH) for a 100 ft^2 (9.3 m^2) space with a 10 ft (3.0 m) ceiling and an airflow rate of 200 cfm (95

kakasveta [241]

Answer:

The ACH is 12/h

Solution:

As per the question:

Area of the space, $A_{s} = 100 ft^{2}$

Height of the given space, h = 10 ft

Air flow rate, $Q_{a} = 200 cfm$

Now, to find the Air Change Rate (ACH):

We calculate the Volume of the given space:

$V_{s} = A_{s}\times h = 100\times 10 = 1000 ft^{3}$

Now, the ACH per min:

= $\frac{V_{s}}{Q_{a}} = \frac{1000}{200} = 5/min$

Now, ACH per hour:

= $\frac{60}{5} = 12/h$

3 0

4 years ago

The two shafts of a Hooke’s coupling have their axes inclined at 20°.The shaft A revolves at a uniform speed of 1000 rpm. The sh

lapo4ka [179]

Answer:

33.429 N-m

Explanation:

Given :

Inclination angle of two shaft, α = 20°

Speed of shaft A, $N_{A}$ = 1000 rpm

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

= 0.1 m

Now we know that for maximum velocity,

$\frac{N_{B}}{N_{A}} = \frac{cos\alpha }{1 - sin^{2}\alpha }$

$\frac{N_{B}}{1000} = \frac{cos20}{1 - sin^{2}20 }$

$N_{B}$ = 1064.1 rpm

Now we know

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

= 0.1 m

Therefore moment of inertia of flywheel, I = m. $k^{2}$

=30 X $0.1^{2}$

= 0.3 kg- $m^{2}$

Now torque on the output shaft

T₂ = I x ω

= 0.3 X 1064.2 rpm

= $0.3\times \frac{2\pi \times 1064.1}{60}$

= 33.429 N-m

Torque on the Shaft B is 33.429 N-m

4 0

3 years ago

The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k

Vesnalui [34]

<h3><u>CSMA/CD Protocol: </u></h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

<u>Example: </u>

$T_{P}=1 H r$ , at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

$\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}$

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

In the previous example we have seen that in worst case $2 T_{P}$ time require to detect a collision.

There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

$\text { Propagation day }=\frac{\text {distance}}{\text {speed}}$

Distance = 1km = 1000m

$\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}$

7 0

3 years ago

The metal control joints used to relieve stresses caused by expansion and contraction in large ceiling or wall expenses in inter

timurjin [86]

Answer:

It’s called Expansion Joints

6 0

3 years ago