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schepotkina [342]
3 years ago
15

A bear spies some honey and takes off from rest, accelerating at a rate of 2.0 m/s^2. If the honey is 16m away, how fast will hi

s snout be going at the moment of ecstasy?
Physics
1 answer:
Anna11 [10]3 years ago
3 0

Answer:

8 m/s

Explanation:

From the third law of motion

v^{2}=u^{2}+2as where v is final velocity, u is initial velocity, a is acceleration, s is distance covered.

Since the bee takes off from rest, initial velocity is zero.

Substituting 0 for u, the equation becomes

v^{2}=0+2as=2as and making v the subject of the formula

v=\sqrt {2as}

Substituting 2 for a and 16 m for s

v=\sqrt {2*2*16}

v=8 m/s

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KIIS FM broadcasts at 102.7 MHz, what is the wavelength of that radio wave?
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3 0
3 years ago
A chevrolet corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. what
pychu [463]
By definition we have that the final speed is:
 Vf² = Vo² + 2 * a * d
 Where,
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 a: acceleration
 d: distance.
 We cleared this expression the acceleration:
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 a = -77268 mi / h ^ 2
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
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 A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
 A = -22.07 feet / sec ^ 2
 Clearing the braking distance:
 Vf² = Vo² + 2 * a * d
 d = (Vf²-Vo²) / (2 * a)
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 answer:
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5 0
3 years ago
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Hope this helps
7 0
3 years ago
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tino4ka555 [31]

Answer:

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