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2 years ago

9

The formation of faults in Earth's crust is an effect. What causes faults to form in the crust? Global Positioning System sensor

s beds shaking across New York City the movement of tectonic plates landslides across the Midwest

1 answer:

allsm [11]2 years ago

7 0

Answer: The movement of tectonic plates

Explanation:

Tectonic plates are the part of the earth's crust that both the ocean and land rest on. These plates are constantly moving as a result of currents in the mantle.

These movements cause stress on the surface which has the effect of fracturing rocks and thereby creating/ forming faults in the earth's crust. Sometimes faults form when these plates move away from each other and sometimes they are formed when they push into each other.

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Given a matrix, clockwise-rotate elements in it. Please add code to problem3.cpp and the makefile. Use the code in p3 to test yo

rusak2 [61]

Answer:

/* C Program to rotate matrix by 90 degrees */

#include<stdio.h>

int main()

{

int matrix[100][100];

int m,n,i,j;

printf("Enter row and columns of matrix: ");

scanf("%d%d",&m,&n);

/* Enter m*n array elements */

printf("Enter matrix elements: \n");

for(i=0;i<m;i++)

{

for(j=0;j<n;j++)

{

scanf("%d",&matrix[i][j]);

}

}

/* matrix after the 90 degrees rotation */

printf("Matrix after 90 degrees roration \n");

for(i=0;i<n;i++)

{

for(j=m-1;j>=0;j--)

{

printf("%d ",matrix[j][i]);

}

printf("\n");

}

return 0;

}

5 0

3 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

2 years ago

What is the biggest expectation when engineers test out designs?

forsale [732]

The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.

6 0

3 years ago

Read 2 more answers

Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3

viktelen [127]

Answer:

Given that

Mass flow rate ,m=2.3 kg/s

T₁=450 K

P₁=350 KPa

C₁=3 m/s

T₂=300 K

C₂=460 m/s

Cp=1.011 KJ/kg.k

For ideal gas

P V = m R T

P = ρ RT

$\rho_1=\dfrac{P_1}{RT_1}$

$\rho_1=\dfrac{350}{0.287\times 450}$

ρ₁=2.71 kg/m³

mass flow rate

m= ρ₁A₁C₁

2.3 = 2.71 x A₁ x 3

A₁=0.28 m²

Now from first law for open system

$h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}$

For ideal gas

Δh = CpΔT

by putting the values

$1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}$

$Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450$

Q= - 45.49 KJ/kg

Q =- m x 45.49 KW

Q= - 104.67 KW

Negative sign indicates that heat transfer from air to surrounding

4 0

2 years ago

The melting point of Pb (lead) is 327°C, is the processing at 20°C hot working or cold working?

bonufazy [111]

Answer:

Explained

Explanation:

Cold working: It is plastic deformation of material at temperature below recrystallization temperature. whereas hot working is deforming material above the recrystallization temperature.

Given melting point temp of lead is 327° C and lead recrystallizes at about

0.3 to 0.5 times melting temperature which will be higher that 20°C. Hence we can conclude that at 20°C lead will under go cold working only.

6 0

3 years ago