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4 years ago

Occurs between a tire on a vehicle and the roadway when a wall of water separates the tire from

Engineering

1 answer:

MatroZZZ [7]4 years ago

4 0

Answer:

a) Hydroplaning

Explanation:

When a wall of water separates the tire on a vehicle from the roadway then the condition is known as hydroplaning or aquaplaning. Skidding of the vehicle is a possibility that may arise due to hydroplaning. However hydroplaning may not immediately result in vehicle stopping abruptly. So among the given options, Hydroplaning fits the scenario specified in the question.

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_____ braking is when you squeeze the brake pedal until just before the wheels lock, then ease off the pedal, then squeeze again

grin007 [14]

The answer should be d) threshold

6 0

3 years ago

Read 2 more answers

In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images

Zarrin [17]

Answer:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

$A= \frac{r^5 \rho}{t^2}$

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= $\frac{ML^2}{T^2}$ A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as $\frac{L}{T}$

[t]=T represent the time

$[\rho]=\frac{M}{L^3}$ represent the density.

Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

And using algebra properties we can express this like that:

$[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}$

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

$-5y = 1$

$y= -\frac{1}{5}$

And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

3 0

3 years ago

Engineer drawing:<br> How can i draw this? Any simple way?

Anastasy [175]

Make 4 triangles left right up down and they must be connected with no gaps then make more triangles into the triangle about three times for each of them then add rectangles or lines to the drawing

6 0

3 years ago

Consider a NACA 2412 airfoil in a low-speed flow at zero degrees angle of attack and a Reynolds number of 8.9·106 . Calculate th

Fofino [41]

Answer:

a) pressure drag is zero (0)

b) pressure drag is 20%

Explanation:

Ans) Given,

NACA 2412 airfoil

Re = 8.9 x 10^6

We know, for turbulent flow ,drag coefficient, Cdf = 0.074 / Re^0.2

=> Cdf = 0.074 / (8.9 x 10^6)^0.2

=> Cdf = 0.003

For both side of plate, Cd = 2 x 0.003 = 0.006

For zero degree angle of attack for NACA 2412, Cdf = 0.006

Also, Cd = Cdf + Cdp

=> 0.006 = 0.006 + Cdp

=> Cdp = 0

Hence, pressure drag is zero

Now, for zero degree angle of attack for NACA 2412, Cd = 0.0075

Also, Cd = Cdf + Cdp

=> 0.0075 = 0.006 + Cdp

=> Cdp = 0.0075 - 0.006

=> Cdp = 0.0015

Hence, Pressure drag percentage = (Cdp / Cdf) x 100

=> Pressure drag percent = (0.0015/0.0075) x 100 = 20 %

Hence, pressure drag is 20% of pressure drag due to flow seperation Ans) Given,

NACA 2412 airfoil

Re = 8.9 x 10^6

We know, for turbulent flow ,drag coefficient, Cdf = 0.074 / Re^0.2

=> Cdf = 0.074 / (8.9 x 10^6)^0.2

=> Cdf = 0.003

For both side of plate, Cd = 2 x 0.003 = 0.006

For zero degree angle of attack for NACA 2412, Cdf = 0.006

Also, Cd = Cdf + Cdp

=> 0.006 = 0.006 + Cdp

=> Cdp = 0

Hence, pressure drag is zero

Now, for zero degree angle of attack for NACA 2412, Cd = 0.0075

Also, Cd = Cdf + Cdp

=> 0.0075 = 0.006 + Cdp

=> Cdp = 0.0075 - 0.006

=> Cdp = 0.0015

Hence, Pressure drag percentage = (Cdp / Cdf) x 100

=> Pressure drag percent = (0.0015/0.0075) x 100 = 20 %

Hence, pressure drag is 20% of pressure drag due to flow seperation

3 0

3 years ago

A distillation column with a partial reboiler and a total condenser is being used to separate a mixture of benzene, toluene, and

marshall27 [118]

Answer:

See attached pictures.

Explanation:

Hello,

In this case, considering the given information and that the feed is a saturated liquid, the solution is shown on the attached pictures with the proper information and procedure.

Best regards.

7 0

3 years ago