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3 years ago

7

Holden is trying to determine the velocity of his race car. He went 20 meters east, turned around, and went 40 meters west. He t

urned the car one more time and went 35 meters east. His car was 15 meters from the starting line. This took 5 seconds. What is the car’s velocity?

2 answers:

AleksandrR [38]3 years ago

6 0

Answer:

3 m /s

Explanation:

Average velocity is defined as the ratio of total displacement to the total time taken. The formula of average velocity is given by

$Average velocity = \frac{Total displacement}{Total time taken}$

Take east direction is positive and west is negative.

Here the total displacement is

d = 20 - 40 + 35 = 15 m

t = 5 second

Average velocity = 15 / 5 = 3 m/s

kotegsom [21]3 years ago

3 0

(15 meters east) / (5 sec) = 3 meters/ sec east

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2 years ago

A Brayton cycle has air into the compressor at 95 kPa, 290 K, and has an efficiency of 50%. The exhaust temperature is 675 K. Fi

motikmotik

Answer:

The specific heat addition is 773.1 kJ/kg

Explanation:

from table A.5 we get the properties of air:

k=specific heat ratio=1.4

cp=specific heat at constant pressure=1.004 kJ/kg*K

We calculate the pressure range of the Brayton cycle, as follows

n=1-(1/(P2/P1)^(k-1)/k))

where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:

P2/P1=(1/0.5)^(1.4/0.4)=11.31

the temperature of the air at state 2 is equal to:

P2/P1=(T2/T1)^(k/k-1)

where T1 is the temperature of the air enters the compressor. Clearing T2

11.31=(T2/290)^(1.4/(1.4-1))

T2=580K

The temperature of the air at state 3 is equal to:

P2/P1=(T3/T4)^(k/(k-1))

11.31=(T3/675)^(1.4/(1.4-1))

T3=1350K

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3 years ago

A cast-iron flywheel has a rim whose OD is 1 m and whose ID is 0.8 m. The flywheel weight is to be such that an energy fluctuati

grin007 [14]

Answer:

A.Coefficient of speed fluctuation of the flywheel = 0.222

B. The width <em>(thickness)</em> of the rim should be 0.131 m (131 mm)

Explanation:

A.

Coefficient of speed fluctuation $(C_{s})$ $= \frac{N_{2}-N_{1}}{N}$

$N_{1}$ = minimum speed = 200 rpm

$N_{2}$ = maximum speed = 250 rpm

$N$ = average speed $= \frac{N_{2}+N_{1}}{2} = \frac{250+200}{2} = 225 rpm$

∴ $Cs = \frac{250-200}{225}=0.222$

Hence the coefficient of speed fluctuation of the flywheel = 0.222

B.

The moment of Inertia , $I=\frac{E_{2}-E_{1}}{C_{s}\times\omega^{2}}$

Where

$E_{2}-E_{1}=$ energy fluctuation of flywheel = 6.75 J

$\omega^{2}=$ angular velocity of flywheel = $\frac{2\pi N}{60} = \frac{2\pi \times 225}{60}= 23.56 rad/sec$

$C_{s}=$ coefficient of speed fluctuation of the flywheel = 0.222

Hence,

$I=\frac{6.75\times10^{3}}{0.222\times(23.56)^{2}}=54.78 Nms^{2}$

Similarly,

I = $\frac{m}{8}\times(d_{o}^{2}- d_{i}^{2})$

From the moment of Inertia, we can get the weight of the flywheel as

$m=\frac{8I}{(d_{o}^{2}+ d_{i}^{2})}= \frac{8\times 54.78}{(1^{2}+0.8^{2})}=267.22kg$

From this weight, we will be able to calculate the volume of the flywheel and hence, estimate the thickness. But to do this, we need to know its density first. this can be got from standard tables.

Specific weight of cast iron = $70.6KN/m^{3}$ ( from standard material property table)

density of cast iron , $\rho =\frac{70.6\times 10^{3}}{9.81}= 7,197kg/m^{3}$

Volume of cast iron flywheel = $\frac{m}{\rho}= \frac{267.22}{7197}= 0.03713 m^{3}$

similarly, the volume of the flywheel can also be obtained through the formula :

$V= \frac{\pi t(d_{o}^{2}-d_{i}^{2})}{4}$

we can easily estimate the thickness of the flywheel from here by solving for t as shown below

$0.03713=\frac{\pi t(1^{2}-0.8^{2})}{4}$

$0.03713=0.2827t$

$t=\frac{0.03713}{0.2827}$

$\therefore t= 0.131m \approx 131mm$

The width of the rim = 131 mm

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4 years ago