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Masja [62]
3 years ago
6

Because of changes over time, the most accurate weather forecasts are: A) analog forecasts. B) long-term forecasts. C) seven-day

s forecasts. D) short-term forecasts.
Physics
2 answers:
blondinia [14]3 years ago
8 0
The answer is analog forecasts
Solnce55 [7]3 years ago
7 0
Because of changes over time, the most accurate weather forecasts are : D. short-term forecasts

average short term forecasts usually only valid for about 36 hours

hope this helps
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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

A = 0.0744m

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

Now the velocity is described as,

v = \frac{2\pi}{T} * \sqrt{A^2-x^2}

v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

v = 0.2049m/s

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3 years ago
ANSWER THIS QUESTION!!!! PLSSSSS!!! NEED HELP!!!!! 10 POINTS
Olegator [25]

Answer:

A. it's the only answer that makes sense. if I'm wrong sorry

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Explanation:

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Answer:

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Explanation:

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