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3 years ago

Verify if 83 is a term in the arithmetic sequence 11,15,19,23

Engineering

1 answer:

EleoNora [17]3 years ago

3 0

Answer:

yes, it is

Explanation:

The sequence: (+4)

23,27,31,35,39,43,47,51,55,59,63,67,71,75,79,83

Hope this helps! :)

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Four eight-ohm speakers are connected in parallel to an audio power amplifier. The amplifier can supply a maximum driver output

Vera_Pavlovna [14]

Answer:

112.5 watts

Explanation:

The output voltage (V) = 15 volts

The equivalent resistance for the speakers connected in parallel ( $R_T$ ) is gotten by using the formula:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\\\But\ R_1=R_2=R_3=R_4=8\ ohm\\\\\frac{1}{R_T} =\frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} \\\\\frac{1}{R_T} =\frac{1+1+1+1}{8} \\\\\frac{1}{R_T} =\frac{4}{8} \\\\R_T=\frac{8}{4} \\\\R_T=2\ ohm$

The current flowing through the amplifier (I) is:

I = V / $R_T$ = 15 / 2 = 7.5 A

The audio power (P) when outputting this maximum voltage is given by:

P = V² / $R_T$ = I² $R_T$ . Therefore:

P = V² / $R_T$ = 15² / $R_T$ = 112.5 watts

6 0

3 years ago

Question is written in following attached screenshot.

Juli2301 [7.4K]

Answer:

jjj jk to the day of the day of the day of the day of the day

Explanation:

hey hi I am a br of August and I am not able to join the meeting and will

7 0

2 years ago

A fatigue test is performed on 69 rotating specimens made of 5160H steel. The measured number of cycles to failure (L in kcycles

tensa zangetsu [6.8K]

Answer:

(a) Mean = 122.9, σ = 30.071

(b) No. of failed specimens at less than 115k cycles are 27.

Explanation:

We are given:

L 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1

(a) First we need to calculate the mean and standard deviation. The formula for calculating mean is:

Mean = ∑fx/∑f

And for standard deviation we have:

S.D. = √Var

Var = ∑fx²/∑f - (Mean)²

∑fx = (2*60) + (1*70) + (3*80) + (5*90) + (8*100) + (12*110) + (6*120) + (10*130) + (8*140) + (5*150) + (2*160) + (3*170) + (2*180) + (1*190) + (0*200) + (1*210)

= 120 + 70 + 240 + 450 + 800 + 1320 + 720 + 1300 + 1120 + 750 + 320 + 510 + 360 + 190 + 0 + 210

∑fx = 8480

Mean = ∑fx/∑f

= 8480/69

Mean = 122.9

∑fx² = (2*60²) + (1*70²) + (3*80²) + (5*90²) + (8*100²) + (12*110²) + (6*120²) + (10*130²) + (8*140²) + (5*150²) + (2*160²) + (3*170²) + (2*180²) + (1*190²) + (0*200²) + (1*210²)

=7200+4900+19200+40500+80000+145200+86400+169000+156800+112500+51200+86700+64800+36100+0+44100

∑fx² = 1104600

Var = ∑fx²/∑f - (Mean)²

= 1104600/69 - (122.9)²

= 16008.69565 - 15104.41

Var = 904.2856

S.D = √Var

σ = √904.2856

σ = 30.071

(b) Let X be the number of failed specimen.

We will use the z-score to calculate the probability. The formula for z-score is:

z = (X-μ)/σ

P(X<115) = P(z<(115-122.9)/30.071)

= P(z<-0.26)

Using the normal distribution probability table, we can compute the value of P(z<-0.26).

P(X<115) = 0.3974

So, no. of failed specimens at less than 115k cycles are: 0.3974*69 = 27 specimens

(c) σ = 30.071

P(x<115) = 0.99

P(z<(115-μ)/30.071) = 0.99

From the normal distribution table we find that 0.99 lies between the z values 2.52 and 2.33. Hence, we get 2.525 as the z-value at which the probability is 0.99.

z = (x-μ)/σ

2.525 = (115 - μ)/30.071

75.93 = 115 - μ

μ = 115 - 75.93

μ = 39.07

4 0

3 years ago

A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a

nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$ = 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$

$T_{1}$ = 20+460 = 480°R

$T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}$ = 533.25°R

Pressure at the inlet of compressor at isentropic condition

$P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}$

$P_{a}$ = $(10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}$ = 14.45 psia

$P_{2}= 8P_{a} = 8(14.45) = 115.6 psia$

4 0

3 years ago

Read 2 more answers

A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is

jonny [76]

Answer:

a) $t = 277.477\,s\,(4.625\min)$ , b) $v_{f} = 0\,\frac{mi}{h}$ , c) $a = -0.128\,\frac{ft}{s^{2}}$

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

$a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}$

$a = -0.185\,\frac{ft}{s^{2}}$

The time required to travel is:

$t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }$

$t = 277.477\,s\,(4.625\min)$

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be $v_{f} = 0\,\frac{mi}{h}$ .

c) The final constant deceleration is:

$a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}$

$a = -0.128\,\frac{ft}{s^{2}}$

7 0

3 years ago