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DerKrebs [107]
2 years ago
9

In the following three scenarios, I do not understand one thing: how do we know that the initial velocity of the y component (vy

0) is zero?
scenario A : Small-plane pilots regularly compete in “message drop” competitions, dropping heavy weights (for which air resistance can be ignored) from their low-flying planes and scoring points for having the weights land close to a target. A plane 60 m above the ground is flying directly toward a target at 45 m/s

scenario B: Trained dolphins are capable of a vertical leap of 7.0 m straight up from the surface of the water—an impressive feat. Suppose you could train a dolphin to launch itself out of the water at this same speed but at an angle. What maximum horizontal range could the dolphin achieve?

scenario C: A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 100 m above the glacier at a speed of 150 m/s. How far short of the target should it drop the package?
Physics
1 answer:
forsale [732]2 years ago
3 0

The y-component of the initial velocity vector is zero only in scenarios A and C. The weight/package on either plane inherits a non-zero x-component that matches the plane's horizontal velocity, but with respect to the vertical direction the objects are at rest, and dropping them from a given height doesn't confer them an initial vertical velocity. On the other hand, if the object was thrown upward and allowed to fall, or shot downward by a cannon, then the initial vertical velocity would be non-zero.

In scenario B, the dolphin must have some non-zero y-component of velocity in order to launch itself out of the water, because otherwise it would stay at a fixed depth.

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A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

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Answer:

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