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2 years ago

An arrow is shot parallel to horizontal from a height of 1.4 meters. It has a velocity of 49 m/s.

Physics

1 answer:

Mkey [24]2 years ago

5 0

The time taken for the arrow to hit the ground is 0.53 s.

The horizontal distance traveled by the arrow is 25.97 m.

The given parameters:

Initial horizontal velocity, $v_x$ = 49 m/s
Initial vertical velocity, $v_y$ = 0

<h3>What is time of motion?</h3>

This is the time taken for a projectile to land on the plane of projection.

The time of motion is calculated as follows;

$h = v_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 1.4}{9.8} }\\\\t = 0.53 \ s$

The horizontal distance traveled by the arrow is calculated as follows;

$X = v_x t\\\\X = (49\ m/s) \times (0.53 \ s)\\\\X = 25.97 \ m$

Learn more about time of motion here: brainly.com/question/2364404

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Vectors A and B lie in the xy ‑plane. Vector A has a magnitude of 17.1 and is at an angle of 150.5∘ counterclockwise from the +x

IRINA_888 [86]

Answer:

A = -14.87 i ^ + 8.42 j ^ + 0 k ^

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Explanation:

For this exercise let's use trigonometry by decomposing to vectors

vector A

module 17.1 with an angle of 150.5 counterclockwise.

Sin 150.5 = $A_{y}$ / A

cos 150.5 = Ax / A

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3 years ago

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3 years ago

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Hitman42 [59]

Answer:

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Explanation:

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where μs is the coefficient of static friction, and Fn is the normal force,

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where ω is the angular velocity of the riders, and r the distance to the

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Since Fc is actually Fn, we can replace the right side of (2) in (1), as

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When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

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Cancelling the masses on both sides of (4), we get:

$g = \mu_{smin} * \omega^{2} * r (5)$

Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

$60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)$

Replacing by the givens in (5), we can solve for μsmín, as follows:

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Answer:

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In this exercise you are asked to enter which sentence is correct, let's start by writing Newton's second law.

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the correct answer is: have a component along the direction of motion that remains perpendicular to the direction of motion

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