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damaskus [11]
3 years ago
12

A snail con move at 3cm/s if it’s ke is 300joules what it it’s mass?

Physics
1 answer:
skelet666 [1.2K]3 years ago
3 0

Explanation:

V = 3 cm/s = 0.03 m/s. BY THE FORmULA OF K.E. K.E = 1/2 mV^2. 300 =1/2 m (0.03)^2. m = 300 x 2/0.0009.

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Car crosses the first third of the road with unknown speed and the rest of the road with the speed of 50 km/h. What is the the s
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The qualitative equivalent of external validity is:
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Which situations might cause two observers (A and B) to measure different frequencies for the same vibrating object? Select the
Alex787 [66]

We want to explain why two different observes may measure different frequencies for the same vibrating object.

We will see that the two correct options are:

  • <em>Observer A is stationary and Observer B is moving.</em>
  • <em>Observer A and Observer B are moving at different speeds relative to each other.</em>

<em />

Let's assume that the vibrating object is a guitar string. Thus, the string makes a noise, and from that noise, we can estimate the frequency at which the string vibrates.

Now there appears a really cool effect, called the Doppler Effect. It says that the apparent change of frequency is <u>due to the motion of the observer or the source of the frequency (or both).</u>

For example, if you move towards the vibrating string, the perceived frequency will be larger, and you will hear a "higher" sound.

While if you move away from the string, the opposite happens, and you will hear a "lower" sound.

Then the only thing that impacts in how we perceive the frequency is our velocity relative to the source.

So, why do observers A and B measure different frequencies?

The two correct answers are:

  • <em>Observer A is stationary and Observer B is moving.</em>
  • <em>Observer A and Observer B are moving at different speeds relative to each other.</em>

If you want to learn more, you can read:

brainly.com/question/17107808

6 0
3 years ago
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
Natasha2012 [34]

The work done by \vec F along the given path <em>C</em> from <em>A</em> to <em>B</em> is given by the line integral,

\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r

I assume the path itself is a line segment, which can be parameterized by

\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath

with 0 ≤ <em>t</em> ≤ 1. Then the work performed by <em>F</em> along <em>C</em> is

\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}

7 0
3 years ago
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