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Nostrana [21]
2 years ago
6

Since you've determined that the power supply is a 700W dual rail, what does that make the maximum output power?

Physics
1 answer:
inysia [295]2 years ago
6 0

700 makes the maximum output power.

<u>Explanation:</u>

In physics, power is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time. The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft.

A joule is equal to one Newton-meter, which is the amount of work needed to move a 1 Newton force a distance of 1 meter. When you divide work by time, you get power, measured in units of joules per second. This is also called a Watt. 1 Watt = 1 Joule Sec. This is the formula to calculate output power.

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A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
Stells [14]

Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

5 0
3 years ago
What selective pressures favored an increase in brain size in hominid primates?
Viktor [21]
The colossal brains of people are to a great extent because of one segment the cerebral cortex. The cerebral cortex is the focal point of dialect, rationale, critical thinking, and data preparing. A high bent in these territories would be exceptionally favorable to primitive seekers.
4 0
3 years ago
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A man in a strength competition pulls an 18-wheel truck 3.10 m in 20.5 s. There is a cable that is attached to his body that exe
larisa [96]

Answer:

114.32195122 but Round your answer to three significant figures.) is 114

Explanation:

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3 years ago
Gravitational force of attraction “F” exists between two point masses A and B when a fixed distance separates them. After mass A
DIA [1.3K]

Answer:

the new gravitational force between the two masses is \frac{3}{2} of the original force (third option in the provided list)

Explanation:

Recall the expression for gravitational force : F_g=G\,\frac{m_A*\,m_B2}{d^2}, where m_A and m_B are the point masses, d the distance between them, and G the universal gravitational constant.

I our problem, the distant between the particles stays unchanged, and we need to know what happens with the magnitude of the force as mass A is tripled, and mass B is halved.

Initial force expression: F_i=G\,\frac{m_A\,m_B}{d^2}

Final force expression: F_f=G\,\frac{3*m_A\,m_B/2}{d^2}\\F_f=G\,\frac{m_A\,m_B\,*\,3/2}{d^2}\\F_f=G\,\frac{m_A\,m_B}{d^2}\,*\frac{3}{2} \\F_f=F_i\,*\frac{3}{2}

Where we have recognized the expression for the initial force between the particles, and replaced it with F_i to make the new relation obvious.

Therefore, the new gravitational force between the two masses is \frac{3}{2} of the original force.

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3 years ago
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A gas bottle contains 0.650 mol of gas at 730 mm Hg pressure. If the final pressure is 1.15 atm, how many moles of gas were adde
AlekseyPX

Answer:

0.779 mol

Explanation:

Since the gas is in a bottle, the volume of the gas is constant. Assuming the temperature remains constant as well, then the gas pressure is proportional to the number of moles:

p \propto n

so we can write

\frac{p_1}{n_1}=\frac{p_2}{n_2}

where

p1 = 730 mm Hg = 0.96 atm is the initial pressure

n1 = 0.650 mol is the initial number of moles

p2 = 1.15 atm is the final pressure

n2 is the final number of moles

Solving for n2,

n_2 = n_1 \frac{p_2}{p_1}=(0.650 mol)\frac{1.15 atm}{0.96 atm}=0.779 mol

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3 years ago
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