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Ahat [919]
2 years ago
12

Question is down below need rply fast

Physics
1 answer:
My name is Ann [436]2 years ago
8 0

The X and Y components are as follows;

1. X = 35 * cos 57 = 19. 1m/s; Y = 35 * sin 57 = 29.4 m/s

2. X = 12 * -cos 34  = -10 m/s; Y = 12 * -sin 34 = -6.7 m/s

3. X = 8 * -cos 90  = 0 m/s; Y = 12 -sin 90 = -8 m/s

4. X = 20 * cos 75 = 5. 2m/s; Y = 20 * (-sin 75) =  -19.3 m/s

<h3>What are the horizontal and vertical components of the vectors?</h3>

The horizontal and vertical components of the velocities are given as follows:

  • Horizontal component, X = x cos θ
  • Vertical component, Y = y sin θ

1. 35 m/s at  57° from x-axis

X = 35 * cos 57 = 19. 1m/s

Y = 35 * sin 57 = 29.4 m/s

2. 12m/s at 34° S of W

X = 12 * -cos 34  = -10 m/s

Y = 12 * -sin 34 = -6.7 m/s

3. 8 m/s at South

X = 8 * -cos 90  = 0 m/s

Y = 12 -sin 90 = -8 m/s

4. 20 m/s at  275° from x-axis

X = 20 * cos 75 = 5. 2m/s

Y = 20 * (-sin 75) =  -19.3 m/s

In conclusion, the X and Y components are found by taking cosines and sine of the angles.

Learn more about horizontal and vertical components at: brainly.com/question/26446720

#SPJ1

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Likurg_2 [28]

Answer:

1)t=2.26\: s

2)S=33.9\: m

3)v=26.77\: m/s

4)\alpha=55.92

Explanation:

1)

We can use the following equation:

y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

0=25-0.5*9.81*t^{2}

t=2.26\: s

2)

The equation of the motion in the x-direction is:

v_{ix}=\frac{S}{t}

15=\frac{S}{2.26}

S=33.9\: m

3)

The velocity in the y-direction of the stone will be:

v_{fy}=v_{iy}-gt

v_{fy}=0-(9.81*2.26)

v_{fy}=-22.17\: m/s

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}

v=26.77\: m/s

4)

The angle of this velocity is:

tan(\alpha)=\frac{22.17}{15}

\alpha=tan^{-1}(\frac{22.17}{15})

\alpha=55.92

Then α=55.92° negative from the x-direction.

I hope it helps you!

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Explanation:

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