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Alika [10]
2 years 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]2 years 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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Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  \alpha =2.538 \  rad/s^2

Explanation:

From the question we are told that

  The mass of the wheel is  m =  6.9  kg

   The radius is  r =  0.69 \  m

    The radius of gyration is  k_G = 0.4\  m

    The angle is  \theta = 47^o

    The force which the massless bar is subjected to F = 22.5 \  N

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that  point A is the  center of rotation.

  Generally the moment of  inertia about A is mathematically represented as

     I_a =  I_G + M* r^2

Here I_G is the moment of inertia about G with respect to the radius of gyration  which is mathematically represented as

    I_G =  M *  k_G

=>I_a = k_G*  M + M* r^2

=>I_a =0.4 * 6.9  + 6.9 * 0.69^2

=>I_a =6.045 \  kg \cdot m^2

Generally the torque experienced by the wheel  is mathematically represented as

       \tau =  F *  cos (47)

=>     \tau =  22.5 *  cos (47)

=>     \tau =  15.34 \ kg \cdot m^2 \cdot s^{-2}

Generally this torque is also mathematically represented as

     \tau = I_a * \alpha

=>   15.34  =  6.045 * \alpha

=>   \alpha =2.538 \  rad/s^2

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The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum
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Answer:

Explanation:

Given

Free fall acceleration on mars g_{m}=3.7\ m/s^2

Time Period of pendulum on earth T=1\ s

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