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Bingel [31]
3 years ago
13

A bumblebee is flying to the right when a breeze causes the bee to slow down with a constant leftward acceleration of magnitude

0.50\,\dfrac{\text m}{\text s^2}0.50 s 2 m ​ 0, point, 50, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction. After 2.0\,\text{s}2.0s2, point, 0, start text, s, end text, the bee is moving to the right with a speed of 2.75\,\dfrac{\text m}{\text s}2.75 s m ​ 2, point, 75, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. What was the velocity of the bumblebee right before the breeze?
Physics
1 answer:
lesya692 [45]3 years ago
7 0

Answer:

3.75 m/s

Explanation:

Given:

v = 2.75 m/s

a = -0.50 m/s²

t = 2.0 s

Find: v₀

v = at + v₀

2.75 m/s = (-0.50 m/s²) (2.0 s) + v₀

v₀ = 3.75 m/s

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Here,

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The tension in the horizontal plane will be subject to the action of the weight, therefore

Tcos\theta = mg

Matching both expressions with respect to the tension we will have to

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\frac{\frac{mv^2}{r}}{sin\theta} =  \frac{mg}{cos\theta}

\frac{mv^2}{r} = mg tan\theta

Rearranging to find the velocity we have that

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6 0
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Answer:

Explanation:

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