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dybincka [34]
3 years ago
6

Please help on this one? The mass on the spring in this diagram is in harmonic motion between x = -5cm and x=5cm

Physics
1 answer:
shutvik [7]3 years ago
8 0

c ... max speed zero accel

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A short circuit is an electrical circuit that allows a current to travel along an unintended path. It has no or very low electrical impedances. The opposite of a short circuit

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A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

E_{A}=E_{B}

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

E_{B}=m*g*h+\frac{1}{2}*m*v^{2}

Therefore we will have the following equation:

(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]

The kinetic energy can be easily calculated by means of the kinetic energy equation.

KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]

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In a series circuit, when batteries are added to the circuit the voltage will _____ and the current will _____.
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Bc of the device for each  server

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