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SCORPION-xisa [38]
3 years ago
9

What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? Vie

w Available Hint(s) What effect does dropping the sandbag out of the cart at the equilibrium position have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude.
Physics
2 answers:
ozzi3 years ago
6 0

Answer:

It has no effect on the amplitude.

Explanation:

When the sandbag is dropped, then the cart is at its maximum speed. Dropping the sand bag does not affect the speed instantly, this is because the energy remains within the system after the bag as been dropped. The cart will always return to its equilibrium point with the same amount of kinetic energy, as a result the same maximum speed is maintained.

amm18123 years ago
5 0

Answer:

It decreases the amplitude.

Explanation:

If it is located at the end, its kinetic energy will be equal to zero, so the system as a whole would only have potential energy, which is defined as:

Ep = (1/2) * k * A²

Where A is the amplitude of the oscillation.

According to the energy principle, said energy must remain conserved, therefore, the total energy is equal to:

Etot = (1/2) * k * A²

Since the system is in equilibrium, according to the expression of conservation of energy, the energy in the position located at the end would be equal to the energy in equilibrium. As the sandbag falls, its kinetic energy decreases, which is why the total energy of the system also decreases.

According to the total energy expression, it is directly proportional to the square of the oscillation amplitude. If the total energy is decreased, the amplitude of the oscillation would also decrease.

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jolli1 [7]

Answer:

Em_{f} / Em₀ = 0.30

Explanation:

In this exercise we use the relationship of mechanical energy, kinetic energy and gravitational potential energy

      K = ½ m v²

      U = mgh

We calculate the mechanical energy at two points

Initial. Lower

    Em₀ = K = ½ m v²

Highest finish

    Em_{f}= U = mg and

Let's calculate

    Em₀ = ½ m 3.6 2

    Em₀ = m 6.48

    Em_{f} = m 9.8 0.20

    Em_{f}= m 1.96

The fraction of lost energy is

    Em_{f} / Em₀ = m 1.96 / m 6.48

   Em_{f} / Em₀ = 0.30

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8 0
3 years ago
You operate a 120 w light bulb for 5 minutes. how much energy did you use?
timurjin [86]

Answer:

Energy use by bulb = 36 kJ

Explanation:

Power of bulb = 120 W

Time of working = 5 minutes =5 x 60 = 300 seconds.

Power = Work / Time

Work = Power x Time

          = 120 x 300

          = 36000 J = 36 kJ

So. energy use by bulb = 36 kJ

3 0
3 years ago
When you throw a ball, the work you do to accelerate it equals the kinetic energy the ball gains. If you do twice as much work w
aniked [119]

Answer:

No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be sqrt{2} times larger.

Explanation:

The work done on the ball is equal to the kinetic energy gained by the ball:

W=K

So when the work done doubles, the kinetic energy doubles as well:

2W = 2 K

However, the kinetic energy is given by

K=\frac{1}{2}mv^2

where

m is the mass of the ball

v is its speed

We see that the kinetic energy is proportional to the square of the speed, v^2. We can rewrite the last equation as

v=\sqrt{\frac{2K}{m}}

which also means

v=\sqrt{\frac{2W}{m}}

If the work is doubled,

W'=2W

So the new speed is

v'=\sqrt{\frac{2(2W)}{m}}=\sqrt{2}\sqrt{\frac{2W}{m}}=\sqrt{2} v

So, the speed is \sqrt{2} times larger.

5 0
3 years ago
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stellarik [79]

Answer:

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iogann1982 [59]

Answer:

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