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2 years ago

Which statement describes the relay between minerals and rocks ?

Engineering

1 answer:

belka [17]2 years ago

6 0

Answer:

Explanation:

Minerals and rocks are the same. Aggregates of minerals form rocks. Minerals determine the texture of a rock. Most rocks are made of a single mineral type.

<h3>☆♡hope this helps♡☆</h3>

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A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a

Dafna1 [17]

Answer:

a) $\dot W = 1.062\,kW$

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}$

$COP_{HP} = 15.692$

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

$\dot Q_{H} = 60000\,\frac{kJ}{h}$

The minimum power supplied to the heat pump is:

$\dot W = \frac{\dot Q_{H}}{COP}$

$\dot W = \frac{\left(60000\,\frac{kJ}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)}{15.692}$

$\dot W = 1.062\,kW$

5 0

3 years ago

Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini

Murrr4er [49]

<u>Explanation:</u>

Task 1 time period = 200ms, Task 2 time period = 300ms

Task ticked = $\frac{1000ms}{200ms}= 5$ → 5 times

Task 2 ticked = $\frac{1000ms}{300ms} = 3.33$ → 3 times

At 600 ms → 200ms 200ms 200ms

300ms → $\frac{30ms}{60ms}$

Largest time period = H.C.M of (200ms, 300ms)

= 600ms

4 0

2 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

2 years ago

A heat recovery system (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loop

blagie [28]

Answer:

0.304 L of Freon is needed

Explanation:

Q = mCT

Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J

C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K

T is temperature in the area of Mars = 189 K

m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg

Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3

Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L

7 0

2 years ago

Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat

zlopas [31]

Answer: $10.631\times 10^3\ N/m^2$

Explanation:

Given

Discharge is $Q=12.5\ L$

Diameter of pipe $d=150\ mm$

Distance between two ends of pipe $L=800\ m$

friction factor $f=0.008$

Average velocity is given by

$\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s$

Pressure difference is given by

$\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa$

8 0

2 years ago