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3 years ago

Whats the difference between GeForce GTX 1060 and Geforce GTX 3060? Is there any big changes to FPS and other settings?

Engineering

2 answers:

ZanzabumX [31]3 years ago

7 0

Answer:

uhhhhh, are you kidding? a GTX 3060 is far better than a 1060 ding dong

Explanation:

mariarad [96]3 years ago

4 0

Answer:

i have a new pc with a 3090 but i had a 3060 before and the difference is GOOOOOD

Explanation:

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Several different loads are going to be used with the voltage divider from Part A. If the load resistances are 300 kΩkΩ , 200 kΩ

harina [27]

Answer:

attached below

Explanation:

7 0

3 years ago

Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t

Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

$h=\frac{2\sigma cos(\alpha )}{\rho gd}$

where

$\sigma$ is the surface tension of the liquid

$\alpha$ is the contact angle of the liquid

$\rho$ is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

$\sigma =7.28\times 10^{-2}N/m$

$\alpha =0$

$\rho _{w}=1000kg/m^3$

Applying the values we get

$h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm$

Part b) When the liquid is mercury:

For mercury and glass we have

$\sigma =485.5\times 10^{-3}N/m$

$\alpha =138^o$

$\rho _{w}=13.6\times 10^{3}kg/m^3$

Applying the values we get

$h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm$

The negative sign indicates that there is depression in mercury in the tube.

4 0

3 years ago

g A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. (a) Obtain the yield strength and ultima

nikdorinn [45]

Answer:

A) - Yield strength before operation = 32 kpsi

- Ultimate Strength before operation = 49.5 kpsi

- Yield strength after operation = 61.854 kpsi

- Ultimate Strength after operation = 61.875 kpsi

- Percentage increase of yield strength = 93.29%

- Percentage increase of ultimate strength = 25%

B) ratio before operation = 1.55

Ratio after operation = 1

Explanation:

From online values of the properties of this material, we have;

Yield strength; S_y = 32 kpsi

Ultimate Strength; S_u = 49.5 kpsi

Modulus; m = 0.25

Percentage of cold work; W_c = 0.2

S_o = 90 kpsi

A) Let's calculate the strain(ε) from the formula;

A_o/A = 1/(1 - W_c)

A_o/A = 1/(1 - 0.2)

A_o/A = 1.25

Thus, strain is;

ε = In(A_o/A)

ε = In(1.25)

ε = 0.2231

Yield strength after the cold work operation is;

S'_y = S_o(ε)^(m)

Plugging in the relevant values;

S'_y = 90(0.2231)^(0.25)

S'_y = 61.854 kpsi

Percentage increase of yield strength = S'_y/(S'_y - S_u) × 100% = (61.854 - 32)/32 × 100% = 93.29%

Ultimate strength after the cold work operation is;

S'_u = S_u/(1 - W_c)

S'_u = 49.5/(1 - 0.2)

S'u = 61.875 kpsi

Percentage increase of ultimate strength = S'_u/(S'_u - S_u) × 100% = (61.875 - 49.5)/49.5 × 100% = 25%

B) Ratio of ultimate strength and yield strength before cold work operations is;

S_u/S_y = 49.5/32

S_u/S_y = 1.547

Ratio of ultimate strength and yield strength after cold work operations is;

S'_u/S'_y = 61.875/61.854 = 1

The ratio after the operation is less than before the operation, thus the ductility reduced.

6 0

3 years ago

For the following pairs of sinusoidal time functions, determine which one leads/lags and by how much. (a) ????1(????)=4sin(6π×10

rjkz [21]

Answer:

The question is incomplete, the complete question is given below

"For the following pairs of sinusoidal time functions, determine which one leads/lags and by how much. (a) V1(t) =4sin(6π×10^4t+60°)V and V(t)2=2cos(6π×10^4t−20°)V. (b) V(t)=10cos(400t−75°) V and I(t)=4sin(400t+30°) A.

Answer

A. V2(t) leads V1(t) by 10°

B. I(t) leads V(t) by 15°

Explanation:

First we express the relationship between sine and cosine of a value.

The expression is giving below Cos (wt) =Sin(wt+90)

Hence for the equations above, we write

a. We can v(t) as

V1(t)=4Sin(6π*10^4+90°-30°)

V1(t)=4Cos(6π*10^4-30°)

Comparing to

V2(t)=4Cos(6π*10^4-20°)

Comparing the angle, we notice that V2(t) leads V1(t) by 10°

b. We can write the current wave form as

I(t)=4sin(400t+90°-60°)

I(t)=4Cos(400t-60°)

If we compare with V(t)=10cos(400t−75°)

I.e 4Cos(400t-60°)=10cos(400t−75°)

We can conclude that I(t) leads V(t) by 15°

4 0

3 years ago

One liter of oven-dry soil sampled from the Chorro Creek Ranch field requires 300 g of water to completely saturate it.

torisob [31]

Answer:

porosity=23.07%

x=5.974*10^9cm^3

Explanation:

One liter of oven-dry soil sampled from the Chorro Creek Ranch field requires 300 g of water to completely saturate it.

Calculate (a) its porosity, n, and (b) the volume of water required to saturate the top 20 cm of 1 hectare of the field.

-Porosity of a soil sample can be calculated in five ways but i shall use the formula below

porosity= $\frac{volume of pore}{volume of total} *100%$ %

volume pore volume of sample is equal to water volume used to saturate the soil

total volume=sample volume +pore volume

1L=1000cm^3

density of water=density=mass/volume

1g/cm^3=300g/volume

vol=300cm^3

total volume=300+1000

1300

porosity =(300/1300)*100%

porosity=23.07%

2. Recall that volume=area *height

1 hectare=1*10^9cm^2

the volume of soil sample=1*10^9cm^2*20cm

2*10^10cm^3

$porosity=\frac{x}{x+2*10^{10} } *100%$ %

23%=x/(x+2*10^10)*100%

x=5.974*10^9cm^3

8 0

3 years ago