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WARRIOR [948]
2 years ago
5

A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference,

this resistor will dissipate:
Physics
1 answer:
Stels [109]2 years ago
7 0

Answer:

<h2>0.056 W</h2>

Explanation:

Power = IV

From ohms law we know that

V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}

\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W

<em>The  resistor will dissipate 0.056 Watt</em>

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A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles i
Tresset [83]

Answer:

\theta_1  = 0.400^o

\theta_2   =0.378^o

Explanation:

From the question we are told that

    The  number of slits per cm is  k =  161\  slits\  per\  cm =  161 \  slits\  per\  0.01 m

    The order of the maxima is  n =  1

    The wavelength are  \lambda_1 = 434 nm  =  434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm =  410 *10^{-9} \  m

The  spacing between the slit is mathematically represented as

           d =  \frac{ 0.01}{k}

=>       d =  \frac{ 0.01}{161}

=>         d = 6.211 *10^{-5} \ m

Generally the condition for constructive interference is  

        n\lambda  =  d \ sin \theta

At  \lambda_1

      \theta _1  =  sin^{-1} [ \frac{1  *  434 *10^{-9}}{6.211 *10^{-5}} ]

      \theta_1  = 0.400^o

At  \lambda_2

       \theta _2  =  sin^{-1} [ \frac{1  *  410 *10^{-9}}{6.211 *10^{-5}} ]

       \theta_2   =0.378^o

4 0
3 years ago
Which of the following is not a part of a wave?
Alisiya [41]
B is the correct answer
y=Asin(wt-kx)
A=amplitude
f=frequency
x=wavelength
since refraction is not on the wave formula,then option B is the correct answer
5 0
2 years ago
Read 2 more answers
Who invented the transistor
erma4kov [3.2K]

A transistor is a semiconductor device used to amplify or switch electronic signals and electrical power. It is composed of semiconductor material usually with at least three terminals for connection to an external circuit. A voltage or current applied to one pair of the transistor's terminals changes the current through another pair of terminals. Because the controlled (output) power can be higher than the controlling (input) power, a transistor can amplify a signal. Today, some transistors are packaged individually, but many more are found embedded in integrated circuits.

Some of the earliest work on semiconductor amplifiers emerged from Eastern Europe. In 1922-23 Russian engineer Oleg Losev of the Nizhegorod Radio Laboratory, Leningrad, found that a special mode of operation in a point-contact zincite (ZnO) crystal diode supported signal amplification up to 5 MHz. Although Losev experimented with the material in radio circuits for years, he died in the 1942 Siege of Leningrad and was unable to advocate for his place in history. His work is largely unknown.

Austro-Hungarian physicist, Julius E. Lilienfeld, moved to the US and in 1926 filed a patent for a “Method and Apparatus for Controlling Electric Currents” in which he described a three-electrode amplifying device using copper-sulfide semiconductor material. Lilienfeld is credited with inventing the electrolytic capacitor but there is no evidence that he built a working amplifier. His patent, however, had sufficient resemblance to the later field effect transistor to deny future patent applications for that structure.

<span>German scientists also contributed to this early research. While working at Cambridge University, England in 1934, German electrical engineer and inventor Oskar Heil filed a patent on controlling current flow in a semiconductor via capacitive coupling at an electrode – essentially a field-effect transistor. And in 1938, Robert Pohl and Rudolf Hilsch experimented on potassium-bromide crystals with three electrodes at Gottingen University. They reported amplification of low-frequency (about 1 Hz) signals. None of this research led to any applications but Heil is remembered in audiophile circles today for his air motion transformer used in high fidelity speakers.</span>

4 0
3 years ago
Read 2 more answers
Puzzle 3
Hunter-Best [27]

Answer:

hagcsgdufgeuwuwgsgwhajisydcbeek

Explanation:

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4 0
3 years ago
A 5 kgkg sphere having a charge of ++ 8 μCμC is placed on a scale, which measures its weight in newtons. A second sphere having
Mrac [35]

Answer:

 F_Balance = 46.6 N    ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

          ∑ F = 0

          Fe –W + F_Balance = 0

         F_Balance = - Fe + W

           

The electric force is given by Coulomb's law

          Fe = k q₁ q₂ / r₂

The weight is

          W = mg

Let's replace

           F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

          q₁ = + 8 μC = +8 10⁻⁶ C

          q₂ = - 3 μC = - 3 10⁻⁶ C

          r = 0.3 m = 0.3 m

Let's calculate

         F_Balance = 5 9.8 - 8.99 10⁹  8 10⁻⁶ 3 10⁻⁶ / (0.3)²

         F_Balance = 49 - 2,397

         F_Balance = 46.6 N

This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

          m' = F_Balance / g

          m' = 46.6 /9.8

          m' = 4,755 kg

6 0
3 years ago
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