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3 years ago

10

A capacitor is connected to a power source and begins charging. What causes the capacitor to stop charging while it is still con

nected to the power source?

1 answer:

Yuri [45]3 years ago

8 0

Charging stops when the voltage across the capacitor reaches the voltage of the power source.

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A plane traveling 40 m/s drops a box from a height of 186 m. Where does the box strike the ground relative to the point directly

Arte-miy333 [17]

Answer:the horizontal velocity is 40

Explanation:

8 0

3 years ago

A dog is running at an initial speed of 10 m/s. He covers 50 m in 4 seconds. What is the acceleration of the dog?

nikdorinn [45]

D i hope this helps
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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

PLEASE HELP ME!!!! i will give brainliest to whoever gets it right

Oksanka [162]

Answer:

I believe the answer is B.)

6 0

3 years ago

The coefficient of linear expansion of steel is 12 × 10-6 K-1 . What is the change in length of a 25-m steel bridge span when it

Westkost [7]

Answer:

0.012-m

Explanation:

∆L = α × Lo × (T-To)

α is the coefficient of linear expansion = 12 × 10-6 K-1

Lo = Initial length = 25-m

∆L = Change in length

(T-To) = 40 K

∆L = 12 × 10-6 × 25 × 40

∆L = 0.012-m

4 0

3 years ago