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2 years ago

14

apply the subtle effect blue accent 1 quick style to the text box. it is the second option in the fourth row of the shape quick

styles gallery

1 answer:

miskamm [114]2 years ago

8 0

To apply the subtle effect, go to the format tab, in the Shape Styles group, select Shape Effects, and click an option from the list. Click the Table Tools Design tab. In the Table Styles group, click the More button. Click the third option in the second row of the Light section.

This is known as editing file in using ms. powerpoint.

Ms powerpoint allows you to create a presentation with combination of text, animation, transition, and etc. If you want to edit your slide in powerpoint, you can select the format bar. There are a lot of feature that can be used to modify your presentation. Ms. powerpoint is launched on April 20, 1987.

Learn more about Ms. PowerPoint, here brainly.com/question/10444759

#SPJ4

You might be interested in

1 gallon of benzene having a density of 0.88 g/mL is spilled in 200 feet by 150 foot lake having an average depth of 6 feet. Wha

Irina18 [472]

Answer:

The concentration of benzene in this contaminated lake would be 653549ng/L.

Explanation:

The concentration C of a contaminant over a body of water of volume V can be obtained as:

C=Mass contaminant / V

In this case, the volume of the body of water:

$V=200ft\cdot150ft\cdot6ft=5097.033m^3$

The mass of the contaminant (Benzene):

$m=\delta V_{ben}=0.88\frac{g}{ml} 1gal(us)=3.3311kg$

Therefore:

$C=\frac{m_{ben}}{V_{lake}}=6.53549\cdot10^{-4}kg/m^3=653549ng/l$

4 0

4 years ago

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at -30C by rejecting

chubhunter [2.5K]

Answer:

a) x = 0.4795

b) QL = 5.85 KW

c) COP = 2.33

d) QL_max = 12.72 KW

Explanation:

Solution:-

- Assuming the steady state flow conditions for both fluids R-134a and water.

- The thermodynamic properties remain constant for respective independent intensive properties.

- We will first evaluate the state properties of the R-134a and water.

- Compressor Inlet, ( Saturated Vapor ) - Ideal R-134a vapor cycle

P1 = 60 KPa, Tsat = -36.5°C

T1 = -34°C , h1 = hg = 230.03 KJ/kg

Qin = 450 W - surrounding heat

- Condenser Inlet, ( Super-heated R-134a vapor ):

P2 = 1.2 MPa , Tsat = 46.32°C

T2 = 65°C , h2 = 295.16 KJ/kg

- Condenser Outlet, ( Saturation R-134a point ):

P3 = P2 = 1.2 MPa , Tsat = 46.32°C

T3 = 42°C , h3 = hf = 111.23 KJ/kg

- R-134a is throttled to the pressure of P4 = compressor pressure = P1 = 60 KPa by an "isenthalpic - constant enthalpy pressure reduction" expansion valve.

- Inlet of Evaporator - ( liquid-vapor state )

P4 = P1 = 60 KPa, hf = 3.9 KJ/kg , hfg = 223.9 KJ/kg

h4 = h3 = 111.23 KJ/kg

- The quality ( x ) of the liquid-vapor R-134a at evaporator inlet can be determined:

x4 = ( h4 - hf ) / hfg

x4 = ( 111.23 - 3.9 ) / 223.9

x4 = 0.4795 Answer ( a )

- Water stream at a flow rate flow ( mw ) = 0.25 kg/s is used to take away heat from the R-134a.

- Condenser Inlet, ( Saturated liquid water ):

Ti = 18°C , h = hf = 75.47 KJ/kg

- Condenser Outlet, ( Saturated liquid water ):

To = 26°C , h = hf = 108.94 KJ/kg

- Since the heat of R-134a was exchanged with water in the condenser. The amount of heat added to water (Qh) is equal to amount of heat lost from refrigerant R-134a.

- Apply thermodynamic balance on the R-134a refrigerant in the condenser:

Qh = flow (mr) * [ h2 - h3 ]

Where,

flow ( mr ) : The flow rate of R-134a gas in the refrigeration cycle

flow ( mr ) = Qh / [ h2 - h3 ]

flow ( mr ) = 8.3675 / [ 295.16 - 111.23 ]

flow ( mr ) = 0.0455 kg/s

- The cooling load of the refrigeration cycle ( QL ) is determined from energy balance of the cycle net work input ( Compressor work input ) - "Win" and the amount of heat lost from R-134a in condenser ( Qh ).

- Apply the thermodynamic balance for the compressor:

Win = flow ( mr )*[ h2 - h1 ] - Qin

Win = 0.0455*[ 295.16 - 230.03] KW - 0.45 KW

Win = 2.513 KW

- The cooling load ( QL ) for the refrigeration cycle can now be calculated. Apply thermodynamic balance for the refrigeration cycle:

QL = Qh - Win

QL = 8.3675 - 2.513

QL= 5.85 KW .... Refrigeration Load, Answer ( b )

- The COP of the refrigeration cycle is calculated as the ratio of useful work and total work input required:

COP = QL / Win

COP = 5.85 / 2.513

COP = 2.33 Answer ( c )

- For a compressor to be working at 100% efficiency or ideal then the maximum COP for the refrigeration cycle would be:

COP_max = [ TL ] / [ Th - TL ]

Where,

TL : The absolute temperature of heat sink, refrigerated space

TH : The absolute temperature of heat source, water inlet

COP_max = [ -30+273 ] / [ (18+273) - (-30+273) ]

COP_max = 5.063

- The theoretical ideal refrigeration load ( QL max ) would be:

COP_max = QL_max / Win

QL_max = Win*COP_max

QL_max = 2.513*5.063

QL_max = 12.72 KW Answer ( d )

5 0

4 years ago

When the psychologist simply records the relationship between two variables...

Wewaii [24]

When a psychologist simply records the relationship between two variables without manipulating them, it is called a correlational study.

The observed relationship does not by itself reveal which variable causes the other. This is the directionally problem. Also, the relationship may be due to a third variable controlling both of the observed variables.

8 0

3 years ago

This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons

telo118 [61]

Explanation:

(a)

Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec

Hence, propagation delay, $d_{prop} = m/sec (s)$

(b)

Here, size of the packet is L bits

And the transmission rate of the link is R bps

Hence, the transmission time of the packet, $d_{trans} = L/R$

(c)

As we know, end-to-end delay or total no delay,

$\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}$

$Here, $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec$ $\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }$

(d)

The expression, time $time $t=d_{\text {trans }}$$ means the\at time since transmission started is equal to transmission delay.

As we know, transmission delay is the time taken by host to push out the packet.

Hence, at $time $t=d_{\text {trans }}$$ the last bit of the packet has been pushed out or transmitted.

(e)

If $\ d_{prop} >d_{trans}$

Then, at $time $t=d_{\text {trans }}$$ the bit has been transmitted from host A, but to condition (1), the first bit has not reached B.

(f)

If $\ d_{prop}$

Then, at $time $t=d_{\text {trans }}$$ , the first bit has reached destination on B

Here, $s=2.5 \times 10^{8} \mathrm{sec}$

$\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}$

It's given that $\ d_{prop} =d_{trans}$

Hence,

$\begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}$

5 0

3 years ago

Determine whether or not each of the following signals is periodic.

Sloan [31]

Answer:

a) periodic (N = 1)

b) not periodic

c) not periodic

d) periodic (N = 8)

e) periodic (N = 16)

Explanation:

For function to be a periodic: f(n) = f(n+N)

$a) x[n]=sin(\frac{8\pi}{2}n+1)\\\\sin(\frac{8\pi}{2}n+1)=sin(4\pi n+1)$

It is periodic with fundamental period N = 1

$b) x[n]=cos(\frac{n}{8} -\pi)\\\\\frac{1}{8} N=2\pi k$

N must be integer. So it is nor periodic

$c) x[n]=cos(\frac{\pi}{8} n^2)\\\\cos(\frac{\pi}{8} (n+N)^2)=cos(\frac{\pi}{8} (n^2+N^2+2nN)\\\\N^2 = 16 \:\:or\:\:2nN=16$

Since N is dependent to n. So it is not periodic.

$d) x[n]=cos(\frac{\pi }{2} n) cos(\frac{\pi }{4} n)\\\\x[n] = \frac{1}{2} cos(\frac{3\pi }{4} n) + \frac{1}{2} cos(\frac{\pi }{4} n)\\\\N_1=8\:\:and\:\:N_2=8\\$

So it is periodic with fundamental period N = 8.

$e) x[n]=2cos(\frac{\pi }{4} n)+sin(\frac{\pi }{8} n)-2cos(\frac{\pi }{2} n+\frac{\pi }{6} )\\\\N_1=8\:\:and\:\:N_2=16\:\:and\:\:N_3=4$

So it is periodic with N = 16.

3 0

3 years ago