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Alexxx [7]
4 years ago
9

A sling is used to give a stone an initial velocity of 20 at an angle of 30 above the horizontal. The stone travels through the

air and lands a horizontal distance of 32m from where it was released. If the stone returns to the same height from which it was thrown, which of the following claims best describes the motion of the air through the stone's trajectory? There must be a horizontal wind in the direction of the stone's motion, because ignoring air resistance when alculating the horizontal range would yield a value less than 32m There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value less than 32m. (C) There is no horizontal wind, because ignoring air resistance when calculating the horizontal range would yield a value equal to 32m. There must be a horizontal wind in the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32m. There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32m
Physics
1 answer:
Luba_88 [7]4 years ago
7 0

Answer:

Option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Explanation:

Normally, ignoring air resistance, for projectile motion, the range (horizontal distance teavelled) of the motion is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile = 20 m/s

θ = angle above the horizontal at which the projectile was launched = 30°

g = acceleration due to gravity = 9.8 m/s²

R = (30² sin 60°) ÷ 9.8

R = 78.53 m

So, Normally, the stone should travel a horizontal distance of 78.53 m. So, travelling a horizontal distance of 32 m (less than half of what the range should be without air resistance) means that, the motion of the stone was impeded, hence, option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Hope this Helps!!!

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A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.
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Answer:

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, b)    f = 0.851 Hz

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a) Hooke's law is

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         k = F / x

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          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

        w = √ k / m

        w = √ (120 / 4.2)

        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

        f = 0.851 Hz

c) the equation that describes the movement is

        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed ​​is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

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       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

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       x = 0

e) the acceleration is

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       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed ​​is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

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i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

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4 years ago
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