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3 years ago

13

In two grams of element x combine with 4 grams of element y to form compound xy how many grams of element why would combined wit

h 43 grams of x to form the same compound

1 answer:

Bingel [31]3 years ago

5 0

Well it takes twice the amount of Y as it does X to make XY. Therefore it would take 2*43 or 86 grams of X

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In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. The sharp upward

11111nata11111 [884]

Answer:

u = 11.6 m/s

Explanation:

The end of a launch ramp is directed 63° above the horizontal. A skier attains a height of 10.9 m above the end of the ramp.

Maximum height, H = 10.9

Let v is the launch speed of the skier. The maximum height attained by the projectile is given by :

$H=\dfrac{u^2\ sin^2\theta}{g}$

$10.9=\dfrac{u^2\ sin^2(63)}{9.8}$

u = 11.6 m/s

So, the launch speed of the skier is 11.6 m/s. Hence, this is the required solution.

4 0

3 years ago

Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

$\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)$

$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

$\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}$

$\vec{A} \times \vec{B} =(0, -33, 12)$

<h2>B)</h2>

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

so:

$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

$\sqrt{1233} = 2 * area_{triangle}$

$35.114= 2 * area_{triangle}$

$17.55 \ units \ of \ area = area_{triangle}$

5 0

3 years ago

How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

$\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}$

Generally, the equation for Rate of flow of Liquid is mathematically given as

$\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}$

$$

Where dP is pressure difference r is the radius

$\mu$ is the viscosity of water

L is the length of the pipe

$Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}$

$Q=5.2 \mathrm{~mL} / \mathrm{s}$

In $30s the quantity that flows out of the tube

$&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}$

In conclusion, the quantity that flows out of the tube

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

Read more about the flows rate

brainly.com/question/27880305

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5 0

2 years ago

I need help ASAP I need to get this right plz plz plz!!!!!

Rina8888 [55]

<em><u>The</u></em><em><u> </u></em><em><u>atomic</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>consists</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>protons</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u>.</u></em>

<em><u>Additional</u></em><em><u> </u></em><em><u>information</u></em><em><u>:</u></em>

<em><u>Protons</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>positive</u></em><em><u>ly</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particl</u></em><em><u>e</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>negative</u></em><em><u>ly</u></em><em><u> </u></em><em><u>charged</u></em><em><u> </u></em><em><u>particle</u></em><em><u>.</u></em>

<em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>will</u></em><em><u> </u></em><em><u>help</u></em><em><u> </u></em><em><u>u</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>:</u></em><em><u>)</u></em>

3 0

3 years ago

Read 2 more answers

(TCO 5) The relationship between Celsius (º C) and Fahrenheit (º F) degree of measuring temperature is linear. Find the linear e

gtnhenbr [62]

Answer:

$C=-\dfrac{10}{17}(F-32)$

Explanation:

Given that

32° F corresponds to 0 °C. ---Point 1

212° F corresponds to 100 °C.----Point 2

We know that if two point is given that equation of line can be found as

$y-y_1=\dfrac{y_2-y_1}{x_2-x_2}(x-x_1)$

Lets C in y- direction and F in x- direction,so we can say that

$C-C_1=\dfrac{C_2-C_1}{F_2-F_2}(F-F_1)$

$C-0=\dfrac{100-0}{32-212}(F-32)$

$C=-\dfrac{10}{17}(F-32)$

So the linear relationship is

$C=-\dfrac{10}{17}(F-32)$

8 0

3 years ago