Measure of how high or low a sound is

A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m

MissTica

Answer:

The magnitude of the vector $\vec{C}$ is 107.76 m

Explanation:

To find the components of the vectors we can use:

$\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

$\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )$

$\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )$

$\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )$

$\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )$

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

$(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)$

So, for our vectors:

$\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ , ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )$

$\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )$

To find the magnitude of this vector, we can use the Pythagorean Theorem

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

$|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}$

$|\vec{C}| =107.76 m$

And this is the magnitude we are looking for.

5 0

3 years ago

This is for physical science, need help :/

marusya05 [52]

1). From the frame of reference of a passenger on the airplane looking out of his window, the tree appears to be moving, at roughly 300 miles per hour toward the left of the picture.

2). The SI unit best suited to measuring the height of a building is the meter.

3). 'Displacement' is the straight-line distance and direction from the start-point to the end-point, regardless of the path that was followed to get there.

The ball started out in the child's hand, and it ended up 2 meters away from her in the direction of the wall. So the displacement of the ball from the beginning to the end of the story is: 2 meters toward the wall.

6 0

2 years ago

Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on

lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

4 0

3 years ago

Calculate the pressure exerted on the heel of a boy’s foot if the boy weighs 80 N and he lands on one heel,which has an area of

cricket20 [7]

Pressure at a given surface is given as ratio of normal force and area

so here force due to heel of the shoes is given as 80 N

and the area of the heel is given as 16 cm^2

so we can say