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34kurt
1 year ago
12

The components of a 15 meters per second velocity at an angle of 60 degrees above the horizontal are?

Physics
1 answer:
mafiozo [28]1 year ago
8 0

Answer:

x-component of velocity: 7.5 m/s

y-component of velocity: 13 m/s

Explanation:

This problem is pure trigonometry. Assuming you know trig, there are only a couple of steps to solving this problem. First, split the velocity into components; recall that any vector not directed along an axis has x and y components. Then, remember that sinΘ = opposite/hypotenuse. Applying this to your scenario, you get sin60° = vy/15. Multiplying this out gives you vy=15sin60. Put this into a calculator (make sure it's set to degree mode because the angle in this problem is in degrees) and you should get 12.99, which you can round up to 13 m/s. This is the velocity in the y-direction.

The procedure to find the x-velocity is very similar, but instead of using sine, we will use the cosine of theta. Recall that cosΘ=adjacent/hypotenuse. Once again plugging this scenario's numbers into that, you end up with cos60 = vₓ/15. Multiplying this out gives you vₓ = 15cos60. Once again, plug this into your calculator. 7.5 m/s should be your answer. This is the velocity in the x-direction.

By the way, a quick way to find the components of a vector, whether it's velocity, force, or whatever else, is to use these functions. Generally, if the vector points somewhere that's not along an axis, you can use this rule. The x-component of the vector is equal to hypotenuse*cosΘ and the y-component of the vector is equal to hypotenuse*sinΘ.

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Which object represents a negatively charged particle? which object represents a positively charged molecule? which object repre
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To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>
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6 0
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49. A block is pushed across a horizontal surface with a
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Answer:

(a) 37.5 kg

(b) 4

Explanation:

Force, F = 150 N

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Net force = mass x acceleration

F - friction force = m a

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150 = m (2.53 + 0.15 x 9.8)

m = 37.5 kg

(b) As the block moves with the constant speed so the applied force becomes the friction force.

F = \mu m g \\\\150 = \mu\times 37.5\\\\\mu = 4

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