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3 years ago

A heat engine is a device that uses _____to produce useful work.

Physics

2 answers:

avanturin [10]3 years ago

8 0

Answer:

Heat

Explanation:

MakcuM [25]3 years ago

3 0

A heat engine is a device that uses heat to produce useful work.

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How much work is done if a 15 newton suitcase is lifted 2 m ?

nirvana33 [79]

Answer:

196 J

Explanation:

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3 years ago

Which of the following objects is accelerating?

Schach [20]

Answer: A flower pot falling

Explanation:

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3 years ago

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Ionic compounds form between elements that are metals and<br> elements that are

NARA [144]

Answer:

nonmetals

Explanation:

Ionic compounds generally form between elements that are metals and elements that are nonmetals. For example, the metal calcium (Ca) and the nonmetal chlorine (Cl) form the ionic compound calcium chloride (CaCl2).

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3 years ago

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

An electric field of 1.27 kV/m and a magnetic field of 0.490 T act on a moving electron to produce no net force. If the fields a

lesantik [10]

Answer:

v = 2591.83 m/s

Explanation:

Given that,

The electric field is 1.27 kV/m and the magnetic field is 0.49 T. We need to find the electron's speed if the fields are perpendicular to each other. The magnetic force is balanced by the electric force such that,

$qE=qvB\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.27\times 10^3}{0.49}\\\\v=2591.83\ m/s$

So, the speed of the electron is 2591.83 m/s.

8 0

3 years ago