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3 years ago

10

The pressure forces on a submersed object will be (A)- Tangential to the objects body (B)- Parallel (C)- Normal (D)- None of the

above

1 answer:

hjlf3 years ago

3 0

Answer:

c) normal

Explanation:

The pressure forces on a submersed object will be normal to the surfaces of the object. Hydrostatic pressure cannot apply tangential forces, so only the normal components exist.

Also, it should be noted that in the hydrostatic case (submerged object) all pressures depend linearly of depth. For small object we can approximate this with equal pressures everywhere.

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Dmitrij [34]

As much as the human and physical capital in economy increases, there is a decrease in the marginal gain in economic growth that will diminish.

<u>Explanation:</u>

Low-income countries might have an advantage achieving greater worker productivity and economic growth in the future as their economic growth is faster than the high - income countries.

As much as the human and physical capital in economy increases, there is a decrease in the marginal gain in economic growth that will diminish. And this is called, the laws of diminishing returns.

Secondly, low - income countries find it easier in developing technologies than the high - income technologies especially countries like India and China.

High - income countries put effort in inventing new technologies, whereas low - income countries just improve and improvise their technology.

3 0

3 years ago

Discuss importance of good communication system

meriva

Answer:

Effective communication helps decision makers by gathering and providing the information to the right person on right time. Communication performs as a motivator to the employees by notifying the employees about the job task, process of carrying and how it could be done better.

7 0

2 years ago

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 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu

Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

The characteristic length of the strip is given by;

$L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m$

The Biot number is given as;

$B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952$

$B_i$ < 0.1, thus apply lumped system approximation to determine the constant time for the process;

$\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s$

The time for the heating process is given as;

$t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s$

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

$T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C$

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

5 0

3 years ago

What is 39483048^349374*3948048/3i4u4

Verizon [17]

Answer:

1. $3.81813506×10^2^9$

2. $1.71479428×10^6^5$

3. $9.38483383×10^2^6$

4. $1.150847×10^2^9$

Explanation:

Feel free to give brainliest

Have a great day!

3 0

2 years ago