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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.

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A student wants to investigate the motion of a ball by conducting two different experiments, as shown in Figure 1 and Figure 2 a
Lelu [443]
It equals to G



^the answer
5 0
2 years ago
A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58. The angel of
dlinn [17]

Answer:

r = 71.8⁰

Explanation:

given,

refractive index of the glass 1 = 1.70

refractive index of glass 2 = 1.58

angle of incidence = 62°

angle of refraction =?

using Snell's law

\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}

\dfrac{sin\ 62^0}{sin\ r} = \dfrac{1.58}{1.70}

1.7 ×sin 62 ^0 = 1.58× sin r

sinr = \dfrac{1.7\times sin 62^0}{1.58}

sin r = 0.95

r = sin⁻¹(0.95)

r = 71.8⁰

angle of refraction =r = 71.8⁰

4 0
2 years ago
Read 2 more answers
Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.
dimulka [17.4K]

Answer:

The distance of separation is d = 0.092 \ m

Explanation:

The mass of the each ball is  m= 10 g  =  0.01 \ kg

 The negative charge on each ball is q_1 =q_2=q =  1 \mu C  =  1 *10^{-6} \ C

Now we are told that the lower ball is  restrained from moving this implies that the net force acting on it is  zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

          F =  \frac{kq_1 * q_2}{d}

=>       m* g  =  \frac{kq_1 * q_2}{d}

here k the the coulomb's  constant with a value  k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

So  

      0.01 * 9.8  =  \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}

            d = 0.092 \ m

5 0
2 years ago
A charge of 87.6 pC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 65.2
LiRa [457]

Answer:

60.8 cm²

Explanation:

The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².

σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²

Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.

Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface

So, q = ε₀Ф = σA'

ε₀Ф = σA'

making A' the area of the Gaussian surface the subject of the formula, we have

A' = ε₀Ф/σ

A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²

A' = 81.4568/1.34 × 10⁻⁴ m²

A' = 60.79 × 10⁻⁴ m²

A' ≅ 60.8 cm²

6 0
2 years ago
First find ∮RB⃗ ⋅dl⃗ , the line integral of B⃗ around a loop of radius R located just outside the left capacitor plate. This can
Allisa [31]

Answer:

the expression of current in the loop enclosed to the left of the capacitor plate is

I(t) = \frac{1}{\mu_0}\int B. dL

Explanation:

As we know by Ampere's law that line integral of magnetic field around a closed loop is proportional to the current enclosed in the path

So we will have

\int B. dL = \mu_0 I(t)

so we have

I(t) = \frac{1}{\mu_0}\int B. dL

so above is the expression of current in the loop enclosed to the left of the capacitor plate

5 0
2 years ago
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