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2 years ago

an airplane flying due north at 90. km/h is being blown due west at 50. km/h. what is the resultant velocity of the plane?

Physics

1 answer:

mina [271]2 years ago

6 0

Given:

$v_{1} = 90 \; km/h$

$v_{2} = 50 \; km/h$

We know that the resultant velocity is,

$v = \sqrt{v_{1} ^{2}+v_{2} ^{2} }$

$v = \sqrt{90^{2}+50^{2}}$

$v=\sqrt{8100+2500}$

$v=\sqrt{10600}$

So, the resultant velocity is,

$v=102.96\;km/h$

<h3>Explain Velocity and its types?</h3>

Velocity is a vector physical term that measures how quickly distance changes with respect to time.

Constant velocity: This refers to the speed at which an object moves through space and time equally.

The speed at which an object cuts various displacements at various moments is known as variable velocity.

Instantaneous velocity: This term refers to either the speed at a specific instant or the constant speed throughout a very brief period of time.

Average velocity is the sum of all distances traveled in a given period of time.

Transverse velocity is the linear speed of an item moving on a circular path. Transverse velocity is the amount of distance covered in a unit of time.

To learn more about Velocity, visit:

brainly.com/question/18084516

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3 years ago

The tension in the horizontal towrope pulling a water-skier is 250 N while the skier moves due west a distance of 50 m. How much

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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:

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:

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

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

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

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

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

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

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

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:

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