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vredina [299]
3 years ago
10

A plane flying horizontally at a speed of 50 m/s and at an elevation of 160 m drops a package, and 2.0 s later it drops a second

package. How far apart will the two packages land on the ground if air resistance is negligible?
Physics
1 answer:
Anarel [89]3 years ago
7 0

Answer:

386 m

Explanation:

Let's call the horizontal distance between the point of launch and the point of landing of the first package x\text{ m}.

2 seconds after landing at x, the plane has travelled a horizontal distance of 2\text{ s}\times 50 \text{ m/s} = 100\text{ m}.

At this new point, the second package is launched. Because it is launched under the same conditions as the first, its horizontal distance from its point of launch is also x\text { m}.

To determine x, we determine the time of fall of the package. Being a vertical motion, the drop has an initial velocity, u, of 0 m/s with acceleration, a = g = 9.8 \text{ m/s}{}^2, and a distance, s, of 160 m. We use the equation of motion:

s = ut+\frac{1}{2}at^2

160 = 0\times t + \frac{1}{2}9.8\times t^2

t = \sqrt{\dfrac{160}{4.9}}\text { s}=\dfrac{40}{7}\text { s}

In this time, the package, with a horizontal velocity of 50 m/s, travels the horizontal distance, x, of

x = 50 \text{ m/s}\times \dfrac{40}{7}\text { s} = \dfrac{2000}{7}\text { m} = 286 \text{ m}

The distance apart between both packages is x+100\text{ m} = 286+100\text{ m} = 386\text{ m}

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The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

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Mass of proton = 1.67 \times 10^{-27} kg

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Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

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