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tamaranim1 [39]

3 years ago

9

A 12000 kg boat is moving 4.25 m/s. Its engine pushes 9200 N forward, but the current pushes back at 12,500 N. How much times do

es it take the boat to stop?

1 answer:

Verizon [17]3 years ago

7 0

Answer:

15.5 seconds

Explanation:

Apply Newton's second law:

∑F = ma

-12500 + 9200 = (12000) a

a = -0.275 m/s²

v = at + v₀

0 = (-0.275) t + 4.25

t = 15.5 s

It takes the boat 15.5 seconds to stop.

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Answer:

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Explanation:

Given that,

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We need to calculate the time

Using equation of motion

$h_{2}=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h_{2}}{g}}$

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$t=0.587\ sec$

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Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

Using conservation of energy

$K.E=mg(h_{1}-h_{2})$

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$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

Put the value into the formula

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

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We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

$d =vt$

Where. d = distance

t = time

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Put the value into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

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3 years ago

Fantom [35]

Answer:

$\mu=0.75$

Explanation:

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