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olchik [2.2K]
2 years ago
12

You've decided to build a new gaming computer and are researching which power supply to buy. Which component in a high-end gamin

g computer is likely to draw the most power? What factor in a power supply do you need to consider to make sure this component has enough wattage?\
Physics
1 answer:
Olenka [21]2 years ago
7 0

Answer:

It is usually the GPU and the CPU in that order. The most you really need to consider is looking at the specs of the parts and add up the wattage. Add 100 Watts to it to be safe and it should be fine. Just purchase a PSU with a 100 watt extra leeway for the pc, which is a negligable price difference.

You might be interested in
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the
Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0
2 years ago
Pistons are fitted to two cylindrical chambers connected through a horizontal tube to form a hydraulic system. The piston chambe
Ivenika [448]

Answer:

order   d> a = e> c> b = f

Explanation:

Pascal's law states that a change in pressure is transmitted by a liquid, all points are transmitted regardless of the form

      P₁ = P₂

Using the definition of pressure

      F₁ / A₁ = F₂ / A₂

      F₂ = A₂ /A₁   F₁

Now we can examine the results

a) F1 = 4.0 N A1 = 0.9 m2 A2 = 1.8 m2

     F₂ = 1.8 / 0.9 4

     F₂a = 8 N

b) F1 = 2.0 N A1 = 0.9 m2 A2 = 0.45 m2

    F₂b = 0.45 / 0.9 2

    F₂b = 1 N

c) F1 2.0 N A1 = 1.8 m2 A2 = 3.6 m2

    F₂c = 3.6 / 1.8 2

    F₂c = 4 N

d) F1 = 4.0N A1 = 0.45 m2 A2 = 1.8 m2

    F₂d = 1.8 / 0.45 4.0

    F₂d = 16 m2

e) F1 = 4.0 N A1 = 0.45 m2 A2 = 0.9 m2

   F₂e = 0.9 / 0.45 4

   F₂e = 8 N

f) F1 = 2.0N A1 = 1.8 m2 A2 = 0.9 m2

   F₂f = 0.9 / 1.8 2.0

   F₂f = 1 N

Let's classify the structure from highest to lowest

F₂d> F₂a = F₂e> F₂c> F₂b = F₂f

I mean the combinations are

 d> a = e> c> b = f

6 0
2 years ago
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
2 years ago
An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t
Setler [38]

Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

\omega = positive

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

\alpha = negative

so here correct answer is

B. W is positive and a is negative

8 0
1 year ago
Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibracion si se le somete a una
EleoNora [17]

Answer:

a)  f = 615.2 Hz      b)  f = 307.6 Hz

Explanation:

The speed in a wave on a string is

         v = √ T / μ

also the speed a wave must meet the relationship

          v = λ f

           

Let's use these expressions in our problem, for the initial conditions

            v = √ T₀ /μ

             √ (T₀/ μ) = λ₀ f₀

now it indicates that the tension is doubled

         T = 2T₀

          √ (T /μ) = λ f

          √( 2To /μ) = λ f

         √2  √ T₀ /μ = λ f

we substitute

         √2 (λ₀ f₀) = λ f

if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change

           λ₀ = λ

           f = f₀ √2

           f = 435 √ 2

           f = 615.2 Hz

b) The tension is cut in half

         T = T₀ / 2

         √ (T₀ / 2muy) =  f = λ f

          √ (T₀ / μ)  1 /√2 = λ f

           fo / √2 = f

           f = 435 / √2

           f = 307.6 Hz

Traslate

La velocidad en una onda en una cuerda es

         v = √ T/μ

ademas la velocidad una onda debe cumplir la relación

          v= λ f  

           

Usemos estas expresión en nuestro problema, para las condiciones iniciales

            v= √ To/μ

             √ ( T₀/μ) = λ₀ f₀

ahora nos indica que la tensión se duplica

         T = 2T₀

          √ ( T/μ) = λf

          √ ) 2T₀/μ = λ f

         √ 2 √ T₀/μ = λ f

         

substituimos  

         √2    ( λ₀ f₀)  =  λ f

si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia

                 λ₀ =  λ

           f = f₀ √2

           f = 435 √2

           f = 615,2 Hz

b)  La tension se reduce a la mitad

         T = T₀/2    

         RA ( T₀/2μ)  =  λ  f

          Ra(T₀/μ) 1/ra 2  =  λ f

           fo /√ 2 = f

           f = 435/√2

           f = 307,6 Hz

5 0
1 year ago
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