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Inga [223]
3 years ago
7

As an object moves from point A to point B only two forces act onit: one force is nonconservative and does -30 J of work, the ot

herforce is conservative and does +50 J of work. Between A andB,
a. the kineticenergy of object increases, mechanical energydecreases.
b. the kineticenergy of object decreases, mechanical energydecreases.
c. the kineticenergy of object decreases, mechanical energyincreases.
d. the kineticenergy of object increases, mechanical energyincreases.
e. None of theabove.
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
7 0

To solve this problem we will apply the principles of energy conservation. On the one hand we have that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.

This leads to the direct conclusion that the resulting energy is 20J.

The conservative force is linked to the movement caused by the sum of the two energies, therefore there is an increase in kinetic energy. The decrease in the mechanical energy of the system is directly due to the loss given by the non-conservative force, therefore there is a decrease in mechanical energy.

Therefore the correct answer is A. Kintetic energy increases and mechanical energy decreases.

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When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some poi
eimsori [14]

Answer: The force of kinetic friction is smaller than that of static friction, but F_g  remains the same.

Explanation:

The situation is same as when a book is pushed with an increasing force on a table; When the force is low, book doesn't move, until that under a given force starts moving, and then it goes on movement even if the force decreases a bit.

The physical explanation for this, that friction force adopts any value needed to avoid to move the object, till a limit value is achieved, called static friction force, equal to the normal force times the static friction coefficient.

Once in movement, the kinetic friction coefficient replaces the static one , and  in general is lower than the static one, so the force diminishes.

In the case of the box sliding down the board, the force that tries to move the object down the board, is the component of the weight parallel to the board, that can be showed that being equal to the weight times the sinus of the angle of the board with the horizontal, as follows:

F_g = m g sin θ

When θ increases, F_g does the same, so friction force always has the same magnitude than F_g (but opposite direction) so the box doesn't move, till that θ takes a value that produces a F_g equal to static friction force.

Beyond this limit angle, F_g (remaining the same for a given angle) is greater than the kinetic friction force, and the box slides.

In the limit, when θ=90º, sin θ =1⇒ F_g = mg, so the object is in free fall.

6 0
3 years ago
When the k. E of
Ierofanga [76]

\sqrt{2}Answer:

KE2 = 2 KE1

1/2 M V2^2 = 2 * (1/2 M V1^2)

V2^2 = 2 V1^2

V2 = \sqrt{2} V1

Since momentum = M V  the momentum increases by \sqrt{2}

8 0
3 years ago
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A train travels 250 km westward from Carthage to Johnson City. The train arrives 2.5 hours after it left. What was the average v
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The answer is C 100 km/h west
5 0
3 years ago
All chemical reactions occur at the same rate true or false​
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Answer:

false

Explanation:

7 0
3 years ago
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Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the
Scrat [10]

Answer: 1.8\°

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d=3.4(10)^{-5}m is the width of the slit

\lambda=540 nm=540(10)^{-9}m is the wavelength of the light  

n is an integer different from zero.

Now, the second-order diffraction angle is given when n=2, hence equation (1) becomes:

dsin\theta_{2}=2\lambda (2)

Now we have to find the value of \theta_{2}:

sin\theta_{2}=\frac{2\lambda}{d} (3)

Then:

\theta_{2}=arcsin(\frac{2\lambda}{d})   (4)

\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})   (5)

Finally:

\theta_{2}=1.8\°   (6)

5 0
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