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VLD [36.1K]
1 year ago
5

What are the advantages and disadvantages of a mine heardgear​

Engineering
1 answer:
Irina18 [472]1 year ago
7 0

Answer:

If there is a shaft with headgear, then mining can take place until that depth of the shaft. If it is accessed horizontal Adits, it can mine until the lowest Adit from upwards. If it is accessed decline, the development and mining can continue so long as economic exploitation is possible.

Explanation:

What are the disadvantages of mining headgear? They totally cut off your vision of anything above your head. They are hot, most of the time

You might be interested in
Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in
Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

Processor clock rate = 2 GHz

Execution Time =   ∑  (\frac{Clock cyles}{Clock rate})

Clock cycles can be determined using following formula

Clock cycles = (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

Clock cycles = ( 50 x 10^{6} x 1) + (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

    = \frac{512(10^{6}) }{2(10^{9}) }

 Initial execution Time =  256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

(\frac{Clockcycles}{2} )=   (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

CPI_{FP improved} x No. FP instructions  =  (\frac{Clockcycles}{2} ) -[ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)]

CPI_{FP improved} x 50 x 10^{6}  = ( \frac{512(10)^{6} }{2} ) - [ (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)]

CPI_{FP improved} x 50 x 10^{6}  =  - 206 x 10^{6}

CPI_{FP improved}  = - 206 x 10^{6} / 50 x 10^{6}

CPI_{FP improved} = - 4.12 < 0

3 0
3 years ago
A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang
AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

\omega =\dfrac{d\theta}{dt}

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

V=\sqrt{\dfrac{GM}{R}}

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

V=\sqrt{\dfrac{GM}{R}}

V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}

V=7059.44 m/s

iii)

We know that centripetal fore given as

F=\dfrac{mV^2}{R}

Here given that m= 200 kg

R= 8000 km

so now by putting the values

F=\dfrac{mV^2}{R}

F=\dfrac{200\times 7059.44^2}{8000\times 10^3}

F=1245.8 N

3 0
2 years ago
17
timofeeve [1]
A. The factor by which a machine multiplies a force
6 0
2 years ago
Compare a series circuit powered by six 1.5-volt batteries to a series circuit powered by a single 9-volt battery. Make sure the
lana [24]

Answer:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

Explanation:

We are asked to compare two series circuits having equal number of light bulbs.

1st circuit is powered by 6 batteries each having a voltage of 1.5V

2nd circuit is powered by a single battery having a voltage of 9V.

The six batteries in the 1st circuit can be connected together in series or in parallel.

When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means

Voltage of pack = number of batteries*voltage of each battery

Voltage of pack = 6*1.5

Voltage of pack = 9 volts

But the current remains same in the series connection since there is only path for the current to flow.

On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.

Circuit 1:

In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Circuit 2:

In this circuit, we have 1 battery which provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit just like in circuit 1.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Conclusion:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

3 0
3 years ago
A horse pulls a cart along a road with a force of 550 lbs. If the horse does 2,674,100 ftlbs of work by the time it stops, how f
BartSMP [9]

Answer:

When a horse pull a cart the action is on?

A horse is harnessed to a cart. If the horse tries to pull the cart, the horse must exert a force on the cart. By Newton's third law the cart must then exert an equal and opposite force on the horse. Newton's second law tells us that acceleration is equal to the net force divided by the mass of the system.

Explanation:

5 0
2 years ago
Read 2 more answers
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