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1 year ago

An object has a moving energy of 25 J. If I do 25 J of work on the object, how much energy does it have now?

Physics

1 answer:

Stella [2.4K]1 year ago

8 0

Given data

*An object has moving energy is U = 25 J

*The work done on the object is W = 25 J

As the work is done on the object to increase its velocity, then the energy it has is given as

$E_n=U+W$

Substitute the known values in the above expression as

$\begin{gathered} E_n=(25)+(25) \\ =50\text{ J} \end{gathered}$

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Answer:

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Explanation:

This problem can be solve using the second law of Newton.

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3 years ago

How long will it take to travel 200.000 m [N] traveling 10 m/s [N]?

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Answer: here are 1,000m in a km, so 200km is 200,000m

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2 years ago

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sineoko [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

Acceleration = (70 / 2.5) m/s²

<em>Acceleration = 28 m/s²</em>

That's about 2.9 G's . Jet pilots can endure a lot more than that, but maybe the catapult or the hook on the airplane can't. Let's look a little closer:

F = m A (Newton #2)

The force on the airplane = (18,000 kg) x (28 m/s²)

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That's about 113,000 pounds ! Maybe the part of the airplane that the catapult pushes on can't handle any more force than that. Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds. Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes ! Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

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3 years ago

A piece of plastic with a mass of 15 g is

satela [25.4K]

Answer:

The density of plastic is equal to 0.6 g/mL.

Explanation:

Given that,

The mass of piece of plastic, m = 15 g

It is placed in a graduated cylinder. The water level in the graduated cylinder rises from 30 mL to 55 mL when the plastic is added.

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2 years ago

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rodikova [14]

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

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$\delta L=L*\epsilon$ (2)

Here L is the member extension and δL the change total longitudinal elongation.

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

$\sigma=P/A$