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NARA [144]
3 years ago
7

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.)
Physics
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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Damm [24]

Answer:

a) 38.27      b) 322.5°

c) 126.99    d) 1.17°

e) 62.27     e) 139.6°

Explanation:

First of all we have to convert the coordinates into rectangular coordinates, so:

a=( 43.3 , 25)

b=( -48.3 , -12.94)

c=( 35.36 , -35.36)

Now we can do the math easier (x coordinate with x coordinate, and y coordinate with y coordinate):

1.)  a+b+c=( 30.36 , -23.3) = 38.27 < 322.5°

2.)  a-b+c=( 126.96 , 2.6) = 126.99 < 1.17°

3.)  (a+b) - (c+d)=0   Solving for d:

     d=(a+b) - c = ( -40.36 , 47.42) = 62.27 < 139.6°

5 0
3 years ago
Find the angle between forces of 41 pounds and 68 pounds given a magnitude of 87 pounds for the resultant force. (Hint: Write fo
dedylja [7]

Answer:

\theta = 76.9 degree

Explanation:

As we know that the resultant of two vectors is given as

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here we know that

R = 87 Lb

F_1 = 41 Lb

F_2 = 68 Lb

now we have

87 = \sqrt{41^2 + 68^2 + 2(41)(68)cos\theta

87^2 = 6305 + 5576 cos\theta

\theta = cos^{-1}(\frac{1264}{5576})

\theta = 76.9 degree

7 0
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Answer:

F = 9.82 N

Explanation:

given,

Force x-component = 5.69 N

Force y-component = 8 N

magnitude of force = ?

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F = \sqrt{5.69^2 + 8^2}

F = \sqrt{32.3761 + 64}

F = \sqrt{96.3761}

F = 9.82 N

Hence, the magnitude of force is equal to 9.82 N

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