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3 years ago

15

25 An 8-bit computer uses 2's complement to store negative numbers. The range of numbers that can be

1 answer:

Inga [223]3 years ago

5 0

Answer:

-128 to +127

Explanation:

8-bits = 2⁸ = 256 possible values

Half of those will be positive and the other half will be negative.

So 128 positive number and -128 negative numbers. But we must also include

the value 0, so now the range will be from -128 to 127

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A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 3

aleksandrvk [35]

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to 250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>

<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table " properties of air " contained in your textbook</em>

average convection heat transfer coefficient = 25.04 W/m^2 .k

<u>ii) Determine how long this process has taken </u>

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

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3 years ago

1. How many pieces of 12-1/2" long wood can be cut from a piece of

Pani-rosa [81]

B. If you do 39-3/4 divided by 12-1/2 you get 3.18

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3 years ago

Select the correct answer. Sean is a recent engineering graduate who has joined a new company. Read the profiles of his colleagu

PSYCHO15rus [73]

Answer:

giving advertising to the governor with the state

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3 years ago

A scale model is 4th the size of the pump. Determine the power ratio of the pump and its scale model if the ratio of the heads i

vredina [299]

Given:

size of scale model = 4(size of pump)

power ratio of pump and scale model = 5:1

Solution:

Let the diameter of scale model and pump be $d_{s}$ and $d_{p}$ respectively

and head be $H_{s}$ and $H_{p}$ respectively

Now, power, P is given as a function of head(H) and dischagre(Q)

P = $\rho gQH$ (1)

From eqn (1):

$P \propto QH$

and

$QH \propto \sqrt{H}D^{2}$

So,

$P \propto H^{\frac{3}{2}} D^{2}$

Therefore,

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{1^{2}\times 5^{\frac{3}{2}}}{4^{2}\times 1^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{5\sqrt{5}}{16}$

${P_{s}}:{P_{p}}$ = ${5\sqrt{5}}:{16}$

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3 years ago

A 4 cm diameter sphere of copper is initially at a temperature of 95 °C. It is placed in a very large water bath at time t- 0. T

avanturin [10]

Answer:112.376 s

Explanation:

Given

$T_i=95^{\circ}C$

$T_f=35^{\circ}C$

$T_\infty \left ( ambient\right )=25^{\circ}C$

$h=400 watts/\left ( m^{2}^{\circ}C\right )$

$c=0.385 J/\left ( m^2^{\circ}C\right )$

$\rho =9 gm/cm^{3}$

Using Newton's law of cooling

$\frac{T_i-T_{\infty}}{T-T_{\infty}}$ = $e^{\frac{ht}{\rho L_{c}c}}$

$\frac{95-25}{35-25}$ = $e^{\frac{400\times 3\times 10^{-4}\times t}{9\times 2\times 0.385}}$

7= $e^{1.7316\times 10^{-2}\times t}$

Taking log both side

t=112.376sec

4 0

3 years ago