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Gemiola [76]
3 years ago
8

How long would it take a boat that started at rest to get to a velocity of 56 m/s if it was accelerating at 17 m/s/s?

Physics
1 answer:
noname [10]3 years ago
8 0

Answer:

3.29 s

Explanation:

Using v = u + at where u = initial velocity of boat = 0 m/s since it starts at rest, v = final velocity of boat = 56 m/s, a = acceleration of boat = 17 m/s² and t = time taken to accelerate to 56 m/s.

So, v -= u + at

t = (v - u)/a

substituting the values of the variables, we have

t = (56 m/s - 0 m/s)/17 m/s²

= 56 m/s ÷ 17 m/s²

= 3.29 s

It will take the boat 3.29 s to accelerate to 56 m/s at an acceleration of 17 m/s²

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cricket20 [7]

Use Newton's 2nd law of motion.
Here it is:

                     Force  =  (mass)  x  (acceleration)

                                 = (80 kg) x (-5 m/s²)

                                 =      -400 Newtons .

I called the acceleration negative because the player is slowing down.

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his motion. 
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Describe the importance of conservative forces to conservation of energy.
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A cylinder of mass mm is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the
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Answer:

y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}

Explanation:

Let y₀ be the initial position of the cylinder when the spring is attached and y its position when it is momentarily at rest.From work-kinetic energy principles,  The work done by the spring force + work done by friction + work done by gravity = kinetic energy change of the cylinder

work done by the spring force = ¹/₂k(y₀² - y²)

work done by friction = - f(y - y₀)

work done by gravity = mg(y - y₀)

kinetic energy change of the cylinder = ¹/₂m(v₁² - v₀²)

So ¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(v₁² - v₀²)

Since the cylinder starts at rest, v₀ = 0. Also, when it is momentarily at rest, v₁ = 0

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(0² - 0²)

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = 0

¹/₂ky₀² + fy₀ - mgy₀ -¹/₂ky² - fy + mgy = 0

¹/₂ky₀² + fy₀ - mgy₀ = ¹/₂ky² + fy - mgy

Let y₀ = 0, then the left hand side of the equation equals zero. So,

0 = ¹/₂ky² + fy - mgy

¹/₂ky² + fy - mgy = 0

Using the quadratic formula

y = \frac{-f +/- \sqrt{f^{2} - 4 X\frac{k}{2} X -mg}}{2 X \frac{k}{2} }\\ y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}

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use the formula

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