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WINSTONCH
2 months ago
11

A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa

lling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target?
Physics
1 answer:
kicyunya [3.2K]2 months ago
6 0
The result is 15.67 seconds. Using the first equation of motion, we have: Final Velocity = Initial Velocity + (Acceleration * Time). This translates mathematically to v = u + at. Now, plugging in our values: v = 3, u = 50, and a = -4 (indicating negative acceleration). We find that 3 = 50 + (-4 * t) leads to -47/-4 = t, giving us a time of 15.67 seconds.
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Answer:

The wire length is 32 inches.

Explanation:

According to the question,

The pipe's length = 60 inches

The pipe's circumference = 4 inches

Now,

The wire wraps around the pipe for 8 complete turns.

Now, with 2πr = 4, we find that

r = 2/π

Thus,

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The wire's length can be calculated as,

Length = Number of turns x Circumference of the pipe

So, length of wire = 8 x 4 = 32 inches.

Consequently, the wire length is 32 inches.

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3 months ago
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Total energy associated with a spring:
E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2

When x = 0.5a:
\frac{1}{2} k \frac{a}{2} ^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2 \\ \frac{1}{2} mv^2 = \frac{1}{2} ka^2 - \frac{1}{8} ka^2 = \frac{3}{8} ka^2

The ratio:
\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} = \frac{3}{4}
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An example of a renewable resource could be a car, a house, or a phone

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Two of the types of ultraviolet light, uva and uvb, are both components of sunlight. their wavelengths range from 320 to 400 nm
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In terms of light energy, a higher frequency corresponds to increased energy within the light.

We establish that frequency is essentially the inverse of wavelength:

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f UVB = 0.0034 to 0.0031

Since UVB occupies a higher frequency range, it consequently possesses greater energy than UVA.

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