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Dafna1 [17]
3 years ago
12

What does a resistor in an electrical dp

Physics
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer:

A resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element. In electronic circuits, resistors are used to reduce current flow, adjust signal levels, to divide voltages, bias active elements, and terminate transmission lines, among other uses.

Explanation:

Resistors are common elements of electrical networks and electronic circuits and are ubiquitous in electronic equipment. Practical resistors as discrete components can be composed of various compounds and forms. Resistors are also implemented within integrated circuits.

The electrical function of a resistor is specified by its resistance: common commercial resistors are manufactured over a range of more than nine orders of magnitude. The nominal value of the resistance falls within the manufacturing tolerance, indicated on the component.

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3 years ago
Under specific circumstances, waves will bend around obstacles. This type of bending is called
BartSMP [9]
That type of bending is called "diffraction" of waves.
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3 years ago
What are the three most abundant gases in the dry atmosphere?
Anastaziya [24]

Answer:

Nitrogen, Oxygen, Argon.

Explanation:

The three (3) most abundant gases in the dry atmosphere are"

- Nitrogen

- Oxygen

- Argon

These are not the only components of dry air. Dry atmosphere is made up of:

- 78.09% Nitrogen;

- 20.95% Oxygen;

- 0.93% Argon;

- 0.04% Carbon dioxide;

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6 0
3 years ago
Read 2 more answers
Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele
JulijaS [17]

Answer: 4.86(10)^{-12}m

Explanation:

The Compton Shift \Delta \lambda in wavelength when photons are scattered is given by the following equation:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta) (1)  

Where:  

\lambda'=500 nm=500(10)^{-9} m is the wavelength of the scattered photon

\lambda_{o}  is the wavelength of the incident photon

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}.c}, being h=4.136(10)^{-15}eV.s the Planck constant, m_{e} the mass of the electron and c=3(10)^{8}m/s the speed of light in vacuum.  

\theta=180\° the angle between incident phhoton and the scatered photon.  

\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°)) (2)

\Delta \lambda=4.86(10)^{-12}m (3)  This is the shift in wavelength

5 0
3 years ago
The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared ​(units are in​ meters). If
ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

y=3+0.8x-0.4x^2........(1)

The x component of constant velocity, v_x=5\ m/s

We need to find the resultant velocity at the point (2,3).

Let \dfrac{dx}{dt}=v_x and \dfrac{dy}{dt}=v_y

Differentiating equation (1) wrt t as,

\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}

v_y=0.8\times v_x-0.8x\times v_x

v_y=0.8v_x(1-x)

When x = 2 and v_x=5\ m/s

So,

v_y=0.8\times 5\times (1-2)

v_y=-4\ m/s

Resultant velocity, v=\sqrt{v_x^2+v_y^2}

v=\sqrt{5^2+(-4)^2}

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

8 0
3 years ago
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