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Schach [20]
4 years ago
6

Select Earth in the menu box so that there is now a force of gravity. Now the total energy of the weight/spring/Earth system is

the sum of the kinetic energy, the elastic potential energy, and the gravitational potential energy. When is the kinetic energy a maximum?
Physics
1 answer:
sergiy2304 [10]4 years ago
4 0

Answer:

When the mass is at the equilibrium position. The total potential energy (the gravitational potential energy plus elastic potential energy) is a minimum at the equilibrium position, even though there is some elastic potential energy when the mass is at this location. The kinetic energy is always a maximum when the total potential energy is a minimum (since the total energy is conserved).

Explanation:

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In addition to gravity, what is the other predominant force that affects the motion of a pendulum?
konstantin123 [22]

When we're working problems that involve a pendulum, we almost always IGNORE anything else besides gravity.  

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This question is probably fishing for the answer "<em>friction</em>".

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A scientist is using an objective lens with 40X magnification on his microscope. If the ocular lens magnifies 10X, what is the t
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Answer:

400x

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3 years ago
Landon attends an early childhood program that is located at a community center which also runs an adult care program.
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3 0
3 years ago
Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

8 0
3 years ago
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