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Studentka2010 [4]

3 years ago

11

Car insurance incentives and discounts are available depending on _____. A. school attendance and driver skill B. vehicle type a

nd driver family size C. location and driver employment status D. location, vehicle type, and driving habits

2 answers:

WARRIOR [948]3 years ago

5 0

Answer:D. Location, vehicle type, and driving habits

BaLLatris [955]3 years ago

3 0

D. Location, vehicle type, and driving habits.

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Check the level of motor oil in your engine by ?

Usimov [2.4K]

Answer:

cpct gvxjjxjhdfjokjdzfjiyddzzsjhxf

6 0

2 years ago

An aircraft is flying at 300 mph true airspeed has a 50 mph tailwind. What is its ground speed?

Free_Kalibri [48]

Answer:

304.13 mph

Explanation:

Data provided in the question :

The Speed of the flying aircraft = 300 mph

Tailwind of the true airspeed = 50 mph

Now,

The ground speed will be calculated as:

ground speed = $\sqrt{300^2+50^2}$

or

The ground speed = $\sqrt{92500}$

or

The ground speed = 304.13 mph

Hence, the ground speed is 304.13 mph

8 0

3 years ago

Two identical 3 in. major-diameter power screws (single-threaded) with modified square threads are used to raise and lower a 50-

sp2606 [1]

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

6 0

3 years ago

water flows in a horizontal constant-area pipe; the pipe diameter is 75 mm and the average flow speed is 5 m/s. At the pipe inle

Veronika [31]

Answer:

Head loss = 28.03 m

Explanation:

According to Bernoulli's theorem for fluids we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=Constant$

Applying this between the 2 given points we have

$\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\gamma _{w}}+\frac{V_{2}^{2}}{2g}+z_{2}+h_{l}$

Here $h_{l}$ is the head loss that occurs

$\therefore h_{l}=\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}-\frac{P_{2}}{\gamma _{w}}-\frac{V_{2}^{2}}{2g}-z_{2}$

Since the pipe is horizantal we have $z_{1}-z_{2}=0$

Applying contunity equation between the 2 sections we get

$A_{1}V_{1}=A_{2}V_{2}\\\\\therefore V_{1}=V_{2}(\because A_{1}=A_{2})$

Since the cross sectional area of the both the sections is same thus the speed

is also same

Using this information in the above equation of head loss we obtain

$h_{l}=\frac{1}{\gamma _{w}}(P_{1}-P_{2})$

Applying values we get

$h_{l}=\frac{1}{9810}\times (275\times 10^{3})m\\\\\therefore h_{l}=28.03m$

3 0

4 years ago

In a 1D compression test with double drainage, the pore pressure readings are practically zero after 8 minutes for a clay sample

frosja888 [35]

Answer:

The duration of the consolidation process for the same clay is 32 min

Explanation:

for clay 1:

t1=0

H1=thickness=2 cm

for the clay 2:

t2=?

H2=2 cm

The time factor is equal to:

$T=(\frac{Cv}{d^{2} })t$

where Cv is the coefficient of consolidation

$(\frac{Cvt}{d^{2} })_{1}= (\frac{Cvt}{d^{2} })_{2}$

if Cv is constant, we have:

$(\frac{t1}{(\frac{H1}{2}) ^{2} })_{1}=(\frac{t2}{H2^{2} })_{2}\\\frac{0}{(\frac{2}{2})^{2}) }=\frac{t2}{2^{2} }$

Clearing t2:

t2=32 min

3 0

3 years ago