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SCORPION-xisa [38]
2 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:
ozzi2 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.

amm18122 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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A 3500 N is force is applied to a spring that has a spring of constant of k= 14000 N/m. How far from equilibrium will the spring
Taya2010 [7]

Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

x = \frac{-F}{K}

Now  

x = \frac{-3500}{14000} \\\\

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm

4 0
3 years ago
Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on
qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

  • Δx=760 m
  • V₀=87 m/s
  • t=13.6 s
  • a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}

760=(1183.2)(cos\theta)

cos\theta=\frac{760}{1183.2}

\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}

3 0
3 years ago
At the moment a hot cake is put in a cooler, the difference between the cakes and the cooler's
hram777 [196]

Answer: D(t)= 50(4/5)^t

Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).

If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.

5 0
2 years ago
Read 2 more answers
Marking branliest!
Ivenika [448]

Answer:

D)evaluating a solution

Explanation:

In this scenario, the next logical step would be evaluating a solution. This is because Jasper and Samantha have already identified the problem/need which is that the robot needs to be able to move a 10-gram weight at least 2 meters and turn in a circle. They also designed and implemented a solution because they have already built the robot. Therefore the only step missing is to evaluate and make sure that the robot they built is able to complete the requirements.

7 0
3 years ago
A coin is dropped. What is its displacement after 0.30 s.
Morgarella [4.7K]
Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.
3 0
2 years ago
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