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miv72 [106K]
4 years ago
7

What made computers different from machines that came before them?

Engineering
2 answers:
e-lub [12.9K]4 years ago
4 0

Answer:

tbh idek I u have to go ask some one els son

Lelu [443]4 years ago
3 0
Early computers were only conceived as calculating devices. Since ancient times, simple manual devices like the abacus aided people in doing calculations. Early in the Industrial Revolution, some mechanical devices were built to automate long tedious tasks, such as guiding patterns for looms.
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Where can you find the grade for a piece of property?
Nadya [2.5K]

Answer:

Simply put, the grade or grading around your house is the level of the ground. The ground level and how it's graded is the deciding factor of where storm water will flow.

Explanation:

5 0
3 years ago
All of these are true about steel EXCEPT that:
natta225 [31]

Answer:B) heat tends to strengthen high-strength steel.

Explanation: You are right it is B

3 0
2 years ago
Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to b
coldgirl [10]

Answer:

A) q'_free = 3146.41 W/m²

B) q'_forced = 7521.41 W/m²

Explanation:

We are given;

Free convection coefficient; h_fr = 5 W/m²K

Force convection coefficient; h_forced = 30 W/m²K

Emissivity; ε = 0.95

Temperature of surrounding which is equal to temperature of air; T_s = T_air = 200°C = 473K

Initial temperature; T_i = 25°C = 298K

A) Now, since the convection feature is disabled, the mode of heat transfer associated with this condition is through free convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_free = q'_free convection + q'_radiation

q'_free convection = h_free(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Where, σ is stephan boltzmann constant and has a constant value of 5.67 × 10^(−8) W/m²K⁴

Thus, rewriting;

q'_free = q'_free convection + q'_radiation

We have;

q'_free = [h_free(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_free = [5(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_free = 875 + 2271.41

q'_free = 3146.41 W/m²

B) Now, in this case, since the convection feature is disabled, the mode of heat transfer associated with this condition is through forced convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_forced = q'_forced convection + q'_radiation

Where;

q'_forced convection = h_forced(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Thus, rewriting;

q'_forced = q'_free convection + q'_radiation

We have;

q'_forced = [h_forced(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_forced = [30(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_forced = 5250 + 2271.41

q'_forced = 7521.41 W/m²

7 0
4 years ago
The acceleration due to gravity at sea level is g=9.81 m/s^2. The radius of the earth is 6370 km. The universal gravitational co
solmaris [256]

Answer:

Mass of earth will be M=5.96\times 10^{24}kg

Explanation:

We have given acceleration due to gravity g=9.81m/sec^2

Radius of earth = 6370 km =6370\times 10^3m

Gravitational constant G=6.67\times 10^{-11}Nm^2/kg^2

We know that acceleration due to gravity is given by

g=\frac{GM}{R^2}, here G is gravitational constant, M is mass of earth and R is radius of earth

So 9.81=\frac{6.67\times 10^{-11}\times M}{(6370\times 10^3)^2}

M=5.96\times 10^{24}kg

So mass of earth will be M=5.96\times 10^{24}kg

3 0
3 years ago
What differentiates an athletic trainer from a clinical exercise physiologist?
kozerog [31]

Answer:

It's number 3

Explanation:

3 0
3 years ago
Read 2 more answers
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