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3 years ago

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A ship sets out to sail to a point 154 km due north. An unexpected storm blows the ship to a point 72 km due east of its startin

g point. How far must it now sail to reach its original destination

1 answer:

Norma-Jean [14]3 years ago

8 0

Answer:

<h2>170km</h2>

Explanation:

If a ship sets out to sail to a point 154 km due north and an unexpected storm blows the ship to a point 72 km due east of its starting point, then the ships distance from the original destination can be gotten by finding the displacement of the ship and this can be gotten by using pythagoras theorem.

Let D be the unknown displacement

According to the theorem;

D² = 154² + 72²

D² = 23716 + 5184

D² = 28900

D = √28900

D = 170km

<em>This means that the ship must now sail a distance of 170km for it to reach its original destination.</em>

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Answer:

a

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b

$H = 9.86 \ m$

Explanation:

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$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

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$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

$v^2 = 0 + 2g[H - 0]$

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When $v = 50 \ km/h = \frac{50 *1000}{3600} = 13.9 \ m/s$ we have that

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3 years ago

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

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$u_2$ = Velocity of the person before collision

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$v_2$ = velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

$v_1 = v_2$

$m_1u_1 + m_2u_2 = (m_1+m_2)v_2$

Re-arrenge to find $v_2$ ,

$v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}$

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$v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}$

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<em />

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3 years ago

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By substitution we obtain,

$16= \frac{1}{2}\times 9.8{t}^{2}$

$\implies32=9.8{t}^{2}$

Diving through by 9.8

$\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}$

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sergejj [24]

Answer:

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$T_2' = T_2 * \frac{9}{5} + 32$

=> $T_2' = 39 * \frac{9}{5} + 32$

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3 years ago