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anastassius [24]
3 years ago
15

A fox sees a piece of carrion men thrown from a Hawk's Nest and rushes to snatch it. The best is 14.0m high and the carrion is t

hrown with horizontal velocity of 1.5 m/s. The fox is 7.0m from the base of the tree. What is the magnitude of the fox average velocity of it grabs the carrion in its mouth just as it touches the ground
Physics
1 answer:
Vsevolod [243]3 years ago
8 0
The y-component of the velocity of the carrion is equal to zero. That being said, the time it takes for the carrion to reach the ground (as close as possible to the fox) can be calculated through the equation,

    d = Vot + 0.5gt²

where d is the distance, Vo is initial velocity (in this case, zero), g is the acceleration due to gravity (9.8 m/s²). Substituting the known values,

  14 = 0.5(9.8)(t²)
    t = 1.69 seconds

Since the horizontal component of the velocity is 1.5 m/s, the distance from the base of the tree to the point where the carrion will fall is equal to,
    (1.5 m/s)(1.69 s) = 2.535 m

We add this to the given distance of the fox from the base of the tree to determine the distance of the fox from the carrion.
   total distance = 2.535 m + 7 m = 9.535 m

Given that the time it takes for it to travel would only be 1.69 seconds, the speed would then be,
 
   speed = (9.535 m) / (1.69 s) = 5.64 m/s

<em>ANSWER: speed = 5.64 m/s</em>
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<h3>What is a bar magnet?</h3>

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A bar magnet like every other magnet possesses a magnetic field that points from the north pole to the south pole.

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2 years ago
A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.
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Answer:

a) The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

\vec p = m\cdot \vec v (1)

Where:

\vec p - Vector momentum, measured in kilogram-meters per second.

m - Mass of the particle, measured in kilograms.

\vec v - Vector velocity, measured in meters per second.

If we know that m = 2.93\,kg and \vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right], then the momentum is:

\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]

\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]

The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}} (2)

\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right) (3)

Where:

\|\vec p\| - Magnitude of momentum, measured in kilogram-meters per second.

\theta - Direction of momentum, measured in sexagesimal degrees.

If we know that p_{x} = 8.731\,\frac{kg\cdot m}{s} and p_{y} = -11.661\,\frac{kg\cdot m}{s}, then the magnitude and direction of momentum are, respectively:

\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}

\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}

\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)

\theta \approx 306.823^{\circ}

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

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EX
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Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

<h3>What is the average speed of the train?</h3>

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

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  • Elapsed time = 3 hours and 40 minutes

First we convert miles to meters and Hours minutes to seconds.

221 miles = ( 221 × 1609.344 )m = 355665.024 meters

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Speed = Distance ÷ time

Speed = 355665.024m / 13200s

Speed = 26.944m/s

Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

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