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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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A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c
Marizza181 [45]

Answer:

11 m/s

Explanation:

Draw a free body diagram.  There are two forces acting on the car:

Weigh force mg pulling down

Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

3 0
3 years ago
A block lies on a frictionless floor. a force of 5 n pulls toward the east while a force of 4 n pulls toward the north. what is
sveticcg [70]

Given below the arrangement of loading on the larger boat by two tug boats.

F₁ = 5 N

F₂ = 4 N

Angle between them θ = 90⁰

Resultant between two vectors, F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta }

Substituting

   F = \sqrt{5^2+4^2+2*5*4*cos 90} \\ \\ = 6.403 N

So magnitude of the net force on the block = 6.403 N

3 0
3 years ago
What is numbers 2 and 4? Please help, and many thanks to those who do chose to help.
Elan Coil [88]

Answer:

2. 6 km/h

4. 150 miles/h

5 0
2 years ago
Which of the following is equal to Velocity change per second​
Mademuasel [1]

Acceleration

Explanation:

Acceleration is a physical quantity that expresses the change in the velocity of a body per unit of time.

  Acceleration = \frac{V - U }{T}

V is the initial velocity

U is the final velocity

T is the time

It is has a unit of m/s²

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

8 0
3 years ago
A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms
Aleonysh [2.5K]

Explanation:

It is given that,

Power of EM waves, P = 1800 W

We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.

Intensity,

I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2

The formula that is used to find the rms value of the electric field is as follows :

I=\epsilon_o cE^2_{rms}

c is speed of light and \epsilon_o is permittivity of free space

So,

E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m

Hence, this is the required solution.

4 0
3 years ago
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