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disa [49]
3 years ago
11

As you brake your bicycle, your velocity changes from 20 east m/s to 10 east m/s in 5 seconds.

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
7 0

Answer:

-2 would be your acceleration

Explanation:.

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Which of these metals are magnetic?
Gwar [14]

Answer:

If iron, nickle, or cobalt is an answer choice, those are the three metals that are purely magnetic.

3 0
2 years ago
An object has rotational inertia I. The object,initially at rest, begins to rotate with a constant angularacceleration of magnit
Tresset [83]

Answer:

L = I α t

Explanation:

given,                                          

rotational inertia = I              

initial velocity = ω₀                      

magnitude of acceleration = α      

angular momentum = L          

time = t                              

angular acceleration                                

\alpha = \dfrac{\omega-\omega_0}{t}

\alpha = \dfrac{\omega - 0}{t}

\alpha = \dfrac{\omega}{t}

ω = α t..............(1)

angular momentum                    

L = I ω                    

putting value from equation (1)

L = I α t                          

4 0
3 years ago
Which of the following statement describes a nonpolar molecule
Sati [7]

Answer:

?

Explanation:

what are the statements

5 0
2 years ago
Read 2 more answers
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rusak2 [61]

Answer:

9.6

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7 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
2 years ago
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