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WINSTONCH [101]
1 year ago
7

Help me please. A B C D

Physics
2 answers:
Hitman42 [59]1 year ago
6 0

Answer:

D

Explanation:

The given vector can be broken down into two components (as shown in the figure).

Which results ; vector (s) = 3x + 4y.

Y_Kistochka [10]1 year ago
5 0

I will go to school tomorrow .....is this present tense or past tense or future tense

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At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t
STatiana [176]
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3 years ago
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Plate tectonics is unique to which terrestrial planet?
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3 years ago
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WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of
nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km

Vector B is 48 km south, so:

B_x = 0\\B_y = -48

Finally, vector C:

C_x = -(100) cos 30^{\circ} =-86.6 km\\C_y = (100) sin 30^{\circ} = 50.0 km

Now we add the components along each direction:

R_x = A_x + B_x + C_x = 36.0 + 0 +(-86.6)=-50.6 km\\R_y = A_y+B_y+C_y = 62.4+(-48)+50.0=64.6 km

So, the resultant (which is the distance in a straight line between the starting point and the final point of the motion) is

R=\sqrt{R_x^2+R_y^2}=\sqrt{(-50.6)^2+(64.6)^2}=82.1 km

4 0
2 years ago
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
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