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denis-greek [22]
3 years ago
13

Why does the static electricity in an object not constantly shock us?

Physics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer:Static electricity is the result of an imbalance between negative and positive charges in an object. These charges can build up on the surface of an object until they find a way to be released or discharged.

Explanation:

hope this helped D:

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If the energies sound in the device is microphone what is the energy out
Katen [24]

Answer:

Microphones are a type of transducer - a device which converts energy from one form to another. Microphones convert acoustical energy (sound waves) into electrical energy (the audio signal). Different types of microphone have different ways of converting energy but they all share one thing in common: The diaphragm.

Explanation:

5 0
3 years ago
Is the frictional force the same as the applied force when the net force equals zero?
DerKrebs [107]

Answer:

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.

Explanation:

6 0
3 years ago
Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d
Schach [20]

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

h_{fg} = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

T_{L} = -18.77°C = 254.23 K

At saturation pressure  800 kPa, the saturation temperature is

T_{H} = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, q_{reject} = h_{fg} = 171.82 kJ/kg

We know COP of heat pump

COP = \frac{T_{H}}{T_{H}-T_{L}}

        = \frac{304.31}{304.31-254.23}

         = 6.076

Therefore, Work out put, W = \frac{q_{reject}}{COP}

                                              = 171.82 / 6.076

                                              = 28.27 kJ/kg

8 0
4 years ago
Which best explains how an object at rest deep in space and far from any massive body behaves compared to an object in free fall
7nadin3 [17]

Answer: They behave the same because, according to the principle of equivalence, the laws of physics work the same in all frames of reference.

Explanation:

According to the equivalence principle postulated by Einstein's Theory of General Relativity, acceleration in space and gravity on Earth have the same effects on objects.

To understand it better, regarding to the equivalence principle, Einstein formulated the following:  

A gravitational force and an acceleration in the opposite direction are equivalent, both have indistinguishable effects.  Because the laws of physics must be accomplished in all frames of reference.

Hence, according to general relativity, gravitational force and acceleration in the opposite direction (an object in free fall, for example) have the same effect.  This makes sense if we deal with gravity not as a mysterious atractive force but as a geometric effect of matter on spacetime that causes its deformation.

3 0
3 years ago
Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to
Debora [2.8K]

Answer:

\mu_{k} \approx 0.719

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a

g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a

\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}

\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )

\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)

\mu_{k} \approx 0.719

5 0
4 years ago
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