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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:

You might be interested in

What flight patterns do groups of birds utilize and why?

lara31 [8.8K]

Birds mostly fly in a V because the lead bird cuts al of the wind and has the hardest work and then after a while the lead bird goes to the back and another bird takes his place cutting all the wind for the other birds

7 0

3 years ago

A heat recovery system (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loop

blagie [28]

Answer:

0.304 L of Freon is needed

Explanation:

Q = mCT

Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J

C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K

T is temperature in the area of Mars = 189 K

m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg

Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3

Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L

7 0

2 years ago

(a) Draw the Moore finite state machine (FSM) of an electronic combination lock with a RESET button, two number buttons (0 and 1

Dafna11 [192]

Answer:

Explanation:

There are a total of 6 states and 3 bits in this problem. Whenever the Reset button is pressed, RESET state is called otherwise the state according to the diagram is called. For the combination to be "01011", the input sequence has to be in the same order. If 0 is pressed instead of 1 in state "010", the last state of output ending with 0 will be called and likewise in all the states that follow.

7 0

3 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the

defon

You need to explain it more simple as everyone is clueless

7 0

2 years ago