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3 years ago

Why was Tetris so popular and use evidence from the text ito support your resp

Engineering

1 answer:

Softa [21]3 years ago

4 0

Answer:

because it was a cool game at that time

Explanation: u didnt give us any text

pls mark brainliest

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Water is flowing into the top of an open cylindrical tank (which has a diameter D) at a volume flow rate of Qi and the water flo

deff fn [24]

Answer:

Z = 3 + 0.23t

The water level is rising

Explanation:

Please see attachment for the equation

8 0

3 years ago

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What major advancement in machine tools occurred in the 1970s and what benefits did it provide? describe in your own words.

mixer [17]

Answer:

I'm just a seventh grader

4 0

3 years ago

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A company buys a machine for $12,000, which it agrees to pay for in five equal annual payments, beginning one year after the dat

Yuki888 [10]

Answer:

$7,778.35

Explanation:

At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.

First, calculate the uniform payments A:

A = 12000(A/P, 4%, 5)

= 12000(0.2246) = 2695.2 (from the calculator)

Then take the last three payments as its own cash flow.

To calculate the new P:

P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35

Therefore, the final payment is $7,778.35

4 0

3 years ago

A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging

dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

$\alpha = \frac{k}{\rho c}$

$= \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s$

Temperature at 28 mm distance after t time = = 50 degree C

we know that

$\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})$

$\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]$

$0.909 = erf{\frac{3.8245}{\sqrt{t}}}$

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have

w = 0.08073

$erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]$

$0.08073 = \frac{3.8245}{\sqrt{t}}$

solving fot t we get

t = 2244.3 sec

3 0

3 years ago

An office building is served by an air-cooled chiller currently operating at 115 tons (404.5 kW). The measured chilled water sup

Andrei [34K]

Answer:

B.197 gpm and 12.4 L/s

Explanation:

Given that

Load Q = 404.5 KW

Water inlet temperature= 6.1 °C

Water outlet temperature= 13.9°C

We know that specific heat for water

$C_p=4.187\ \frac{KJ}{kg.K}$

Now from energy balance

$Q=\dot{m}C_p\Delta T$

by putting the values

$Q=\dot{m}C_p\Delta T$

$404.5=\dot{m}\times 4.187(13.9-6.1)$

$\dot{m}=12.38\ \frac{kg}{s}$ (1 Kg/s = 15.85 gal/min)

We can say that

$\dot{m}=196.31\ \frac{gal}{min}$

We know that

$\dot{m}=\rho\times volume\ flow\ rate$

12.38=1000 x volume flow rate

$volume\ flow\ rate\ = 12.38\times 10^{-3}\ \frac{m^3}{s}$

volume flow rate = 12.38 L/s

So the option B is correct.

8 0

3 years ago