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3 years ago

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A cruise ship with a mass of 1.03 ✕ 107 kg strikes a pier at a speed of 0.652 m/s. It comes to rest after traveling 4.44 m, dama

ging the ship, the pier, and the tugboat captain's finances. Calculate the average force (in N) exerted on the pier using the concept of impulse. (Hint: First calculate the time it took to bring the ship to rest, assuming a constant force. Indicate the direction with the sign of your answer.)

1 answer:

Gnoma [55]3 years ago

5 0

Answer:

The average force exerted on the pier is $4.9 \times 10^{5}$ N

Explanation:

Given :

Mass of ship $m = 1.03 \times 10^{7}$ Kg

Distance $x = 4.44$ m

Velocity $v = 0.652$ $\frac{m}{s}$

Now find the average velocity,

$v _{avg} = \frac{0+0.652}{2}$

$v _{avg} = 0.326$ $\frac{m}{s}$

Now find the time taken by ship to travel 4.44 m.

$t = \frac{x}{v_{avg} }$

$t = \frac{4.44}{0.326}$

$t = 13.62$ sec

Now calculating average force,

$F = \frac{m \Delta v}{\Delta t}$

Where $\Delta v = 0.652 -0$ = 0.652 $\frac{m}{s}$

$F = \frac{1.03 \times 10^{7} \times 0.652 }{13.62}$

$F = 4.9 \times 10^{5}$ N

Therefore, the average force exerted on the pier is $4.9 \times 10^{5}$ N

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Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

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and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

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An assault rifle fires an eight-shot burst in 0.40 s. Each bullet has a mass of 7.5 g and a speed of 300 m/s as it leaves the gu

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Answer:

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The change in momentum of an object is equal to impulse delivered. So,

$F\times t=mv\\\\F=\dfrac{mv}{t}$

For 8 shot burst, average recoil force on the gun is :

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Answer:

a) the Angle te also decreases , b) decrease, c) unchanged , d) the speed decrease , e) unchanged

Explanation:

When a ray of light passes from one transparent material to another, it must comply with the law of refraction

n1 sin θ₁ = n2 sin θ₂

Where index v1 is for the incident ray and index 2 for the refracted ray

With this expression let's examine the questions

a) They indicate that the refractive index increases,

sin θ₂ = n₁ / n₂ sin θ₁

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Answer is decrease

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The answer is decrease

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klasskru [66]

Answer:

(A) 7.9 m/s^{2}

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(C) 91 m

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