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3 years ago

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In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

ther step to isolate it. Let's say we now have
3C+4D = 5
2C+5D = 2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?
a. C=53−43D
b. C=1−52D
c. D=25−25C
d. D=54−34C

1 answer:

Flura [38]3 years ago

5 0

Answer:

a) C = (5-4D)/3

Explanation:

Given the simultaneous equation

3C+4D = 5 .... 1

2C+5D = 2 .... 2

In order to isolate one of the variables, we will make one of the variables in any of the equation.

Using equation 1:

3C+4D = 5

Make C the subject of the formula:

Subtract 4D from both sides of the equation.

3C+4D-4D= 5-4D

3C = 5-4D

Divide both sides by 3:

3C/3 = (5-4D)/3

C = (5-4D)/3

Hence the expression that would result from the process is C = (5-4D)/3

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Answer:

The required total area is 1.48 m²

Explanation:

Given that,

Latitude = 44+° N

New Mexico,

Latitude= 35+° N

Heat capacity = 4200 J/Kg°C

Temperature = 60°C

Let us assume the input temperature 22°C

Estimate volume of water 100 ltr for 4 person.

We need to calculate the heat

Using formula of heat

$H=mc_{p}\Delta T$

$H=mc_{p}(T_{f}-T_{i})$

Put the value into the formula

$H=100\times4200\times(60-22)$

$H=15960\ KJ$ ...(I)

Let solar radiation for 6 hours/day.

We need to calculate the total energy per unit area

Using formula of energy

$E=1000\times6\times3600\ J/m^2$

$E=21600\ KJ/m^2$

Let the efficiency of collector is 50 %

Then, the total energy per unit area will be

$E=21600\times\dfrac{50}{100}$

$E=10800\ KJ/m^2$ ....(II)

We need to calculate the required total area

Using equation (I) and (II)

$A=\dfrac{H}{E}$

Where, H = heat

E = total energy

Put the value into the formula

$A=\dfrac{15960}{10800}$

$A=1.48\ m^2$

Hence, The required total area is 1.48 m²

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3 years ago

Jim is driving a 2268-kg pickup truck at 22 m/s and releases his foot from the accelerator pedal. The car eventually stops due t

shutvik [7]

Answer:

610 meters.

Explanation:

Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.

the kinetic energy of the truck just after Jim released the pedal is:

$E_k=\frac{1}{2}*m*v^2\\E_k=\frac{1}{2}*2268*(22)^2=548856J$

The work done by the friction force is given by:

$W_f=F_s*d\\\\d=\frac{548856J}{900N}\\\\d=610m$

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3 years ago

X-rays cannot pass through Earth's atmosphere. Which of these is the best location to place a telescope used to observe x-rays f

Amiraneli [1.4K]

Mountains, tops of buildings, and high-flying aircraft are all part of Earth's atmosphere, no matter how high they are. On the other hand, space doesn't belong to our atmosphere, it is outside of it. Having this in mind, the best location to place a telescope used to observe x-rays from stars is in space.

7 0

3 years ago

Read 2 more answers

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

A helicopter flies with an air speed of 175 km/h, heading south. The wind is blowing at 85 km/h to the east relative to the grou

spayn [35]

Answer:

154° at 195 km/h

Explanation:

The helicopter is moving south at 175 km/h, relative to the wind.

But the wind is moving east at 85 km/h, relative to the ground.

This means that the helicopter is moving south east relative to the ground.

Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.

This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.

Refer to the triangle b1.

The distance traveled by the helicopter in 1 hour is denoted by d.

d is the hypotenuse of the right triangle.

Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)

Hence the helicopter is traveling at 195 km/h relative to the ground.

To calculate the direction we use,

tan (x) = opposite/adjacent = 85/175

So the angle x is,

$x = arctan (\frac{85}{175} )$ = 25.9°

Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)

8 0

3 years ago