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irina1246 [14]
3 years ago
14

The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters

before coming to rest in a lock containing 75,000 cubic meters of fresh water. The specific heat of water is 4200 joules per kilogram degree Celsius. Assume all energy is transferred from the ship to the water. Determine the change in temperature of the water in degrees Celsius .
Physics
1 answer:
skelet666 [1.2K]3 years ago
5 0

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh   (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT   (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C    (3)

ΔT = 5.39x10⁹ / 4200 * 75000

<h2>ΔT = 17.11 °C</h2>

Hope this helps

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3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
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Answer:

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Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

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a=2.25 \mathrm{m} / \mathrm{s}^{2}

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\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

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3 years ago
If a wave had a wave speed of 1000 m/s and a frequency of 500 Hz, what is its wavelength?
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<h2>2 meters</h2>

Explanation:

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Answer:

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