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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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Ken received a 66 on his first math exam, which counted for 20% of his final grade; he now believes that he won't be able to pas
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Answer:

His conclusion best illustrates a pessimistic outlook.

Explanation:

As seen in the question above, Ken got 20% of his final grade in the first test he did for this class, that is, there will be other tests that can provide him to reach the grade needed to pass the class. However, even if there are possibilities, he believes that he will not pass the class, he does not have a positive and optimistic view of his future in this class and is sure that he will fail. This negative view of the future is an example of a pessimistic outlook.

6 0
3 years ago
A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is th
IceJOKER [234]

Answer:E=75\ N/m

Explanation:

Given

First charge of q_1=15\ nC is placed at x=1.5\ m

Second charge  q_2=-20\ nC is placed at y=-2\ m

Electric field is given by

E=\frac{kq}{r^2}

Electric field due to q_1 is away from it

E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}

E_1=60\ N/m

Electric field due to q_2

E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}

E_2=45\ N/m

Net electric field will be vector addition of two

\vec{E_{net}}=\vec{E_1}+\vec{E_2}

\vec{E_{net}}=-60\hat{i}-45\hat{j}

Magnitude of Electric field is

E=\sqrt{60^2+45^2}

E=75\ N/m

8 0
3 years ago
(a) A hole of 1 cm in diameter is drilled in a plate of steel at 20°C. What happens to the diameter of the hole as the steel is
Anestetic [448]

When heat is applied to a substance/material or steel in our case, a series of changes will occur, one of which is Expansion.

Hence the hole will expand from 1cm to a diameter that is greater.

Thermal expansion refers to the expansion or contraction of the dimensions of the solid, liquid or gas when their temperature is changed.

<h3>Types of Expansion</h3>
  • Linear expansion
  • Areal expansion and
  • Volumetric volume.

Learn more about expansion here:

brainly.com/question/1166774

8 0
3 years ago
Which symbol is used to show vector quantities
hjlf

Answer:  arrows

Explanation:

6 0
3 years ago
Question in picture.
Gelneren [198K]
<h2>Hello!</h2>

The answer is: A. 19.3 joules

<h2>Why?</h2>

Since it's an elastic collision, the kinetic energy after and before the collision will be the same.

Kinetic energy can be calculated using the following equation:

KE=\frac{1}{2}mv^{2}

Where:

KE=KineticEnergy\\m=mass\\v=velocity

So,

First object, (going to the right):

m=7.20kg\\v=2\frac{m}{s}

KE_{1}=\frac{1}{2}*7.20Kg*(2\frac{m}{s})^{2}=14.4Joules

Second object:, (going to the left):

m=5.75kg\\v=-1.30\frac{m}{s}

KE_{2}=\frac{1}{2}*5.75kg*(-1.30\frac{m}{s})^{2}=4.86Joules

Remember,

1Joule=1Kg.\frac{m^{2}}{s^{2} }

Hence,

The total kinetic energy after the collision will be:

T=KE_{1}+KE_{2}=14.4Joules+4.86joules=19.26joules=19.3joules

The total kinetic energy after the collision is 19.3 joules (rounded to the nearest tenth)

Have a nice day!

6 0
3 years ago
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