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ICE Princess25 [194]
3 years ago
11

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

ips by calculating their velocity just after they collided. According to the last radio transmission from the 40,000-ton luxury liner, the Hedonist, it was going due west at a speed of 20 knots in calm seas through a rare fog just before it was struck broadside by the 60,000-ton freighter, the Ironhorse, which was traveling directly northeast at 10 knots. The transmission also noted that when the freighter's bow pierced the hull of the liner, the two ships stuck together and sank.To determine the search area, find the final velocity vector just after moment of collision.
Physics
1 answer:
Leya [2.2K]3 years ago
6 0

<u>Solution and Explanation:</u>

The following calculation is made in order to find out the area and the final velocity vector.

Using the given information and the data in the question,  

m1u1 + m2u2 = (m1 + m2)  multiply with v

40000 multiply with ( -20i ) + 60000 multiply with ( 10j ) = 100000 multiply with v

  Therefore, v = -8i + 6j

That is  |v| = 10 knots towards the 36.86 degree north of the west

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<span>The change in the electron's potential energy is equal to the work done on the electron by the electric field. The electron's potential energy is the stored energy relative to the electron's position in the electric field. Vcloud - Vground represents the change in Voltage. This voltage quantity is given to be 3.50 x 10^8 V, with the electron at the lower potential. The formula for calculating the change in the electron's potential energy (EPE) is found by charge x (Vcloud - Vground) = (EPEcloud - EPE ground) where charge is constant = 1.6 x 10^-19. Filling in the known quantities results in the expression 1.6 x 10^-19 (3.50 x 10^8) = (EPEcloud - EPEground) = 5.6 x 10^-11. Therefore, the change in the electron's potential energy from cloud to ground is 5.6 x 10^-11 joules.</span>
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3 years ago
A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac
Tpy6a [65]

Answer:

The force on one side of  the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

y=\dfrac{\Delta y}{\sin60}

y=\dfrac{2\Delta y}{\sqrt{3}}

The area of strip is

dA=\dfrac{9\times2\Delta y}{\sqrt{3}}

We need to calculate the force on  one side of  the plate

Using formula of pressure

P=\dfrac{dF}{dA}

dF=P\times dA

On integrating

\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}

F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}

F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})

F=3093529.3\ N

Hence, The force on one side of  the plate is 3093529.3 N.

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3 years ago
A 0.095 kg tennis ball is traveling to the right at 40 m/s when it bounces of a wall and travels in the opposite direction it ca
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given that

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\Delta P = 0.095(30 - (-40))

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Got shot with a pump shotgun to the head
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Which statements best describe x-rays? check all that apply. x-rays are electromagnetic waves. x-rays are longitudinal waves. x-
igor_vitrenko [27]

X ray is one of the electromagnetic waves.

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3 0
3 years ago
Read 2 more answers
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