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Alika [10]
1 year ago
7

After combining components of two vectors to be added, it was found that the resultant vector has an x-component of -177 cm and

a y.component of -214 cm
What angle measured counterclockwise from the positive x-axis will properly describe the direction of the resultant vector? (Round to the nearest tenth of a
degree.)
0 219,6
O 50.4
оооо
O 3204
0 230.4
Physics
1 answer:
3241004551 [841]1 year ago
8 0

The angle measured counterclockwise from the positive x-axis is θ = 50.4°

<h3>How to get the angle correspondent to a vector?</h3>

Here we know that the vector is:

V = < -177 cm, -214 cm>

To get the correspondent angle for this vector, we can think that this is the hypotenuse of a right triangle, such that the y-component and x-component are the cathetus.

Then, to get the angle (measured counterclockwise from the positive x-axis) is given by:

Tan(θ) = (opposite cathetus)/(adjacent cathetus)

Tan(θ) = (-214cm)/(-177 cm)

Using the inverse tangent function we get:

Atan(Tan(θ)) = Atan((-214cm)/(-177 cm))

θ = 50.4°

So the angle is 50.4°

If you want to learn more about vectors, you can read:

brainly.com/question/3184914

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HELP!!
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Hello!

Since the two weights are <em>off</em> the table, the block will move towards letter F. 

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Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
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The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

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yes

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C = 261.6Hz × 1.31m
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