All categories

2 years ago

True or False Most coolants you put into your car are toxic.

Engineering

2 answers:

swat322 years ago

8 0

Answer: True.

Explanation: Coolant is flammable and toxic.

vaieri [72.5K]2 years ago

5 0

True
Anything you put in your car is toxic

You might be interested in

11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t

lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

$v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2} }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2} }{4}) } =0.729m/s$

It is necessary to get the Reynold's number:

$Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343$

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

$Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517$

The overall heat transfer coefficient:

$Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} } }$

Here

$h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C$

Substituting values:

$Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278} } =1855.8923W/m^{2} C$

5 0

3 years ago

Which of these cars traveled faster during time interval <br> please show solution

Mazyrski [523]

Answer:

I think D is correct

Explanation:

C is decreasing function, probably worst

A is arctan -> in radian, the rate of increasing is very slow-> second worst

B(14) = ln(9*14) = 4.8

D(14) = sqrt(8+14^2)=14.2

3 0

3 years ago

Hi I don't know of yall remeber me, but I'm Jadin aka J. I am looking for my friend group, that I have missed but can't find cau

kirill [66]

Answer:

The young lady was his daughter.The shoemaker was frightened when he saw that she wants to sit near him and took his knife to frighten her and leave him alone to do his work

Explanation:

could uh name them since if i know any i would surely tryin help

7 0

2 years ago

Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

Dafna11 [192]

Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

The initial volume of water $V_1$ = 1000.00 mL = 1000000 mm³

The initial temperature of the water $T_1$ = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water $T_2$ = 70° C

The coefficient of thermal expansion for the glass is ∝ = $3.8*10^{-6 } mm/mm \ per ^oC$

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature $T_1 = 10 ^ 0C$ the density of the water is $\rho = 999.7 \ kg/m^3$

At temperature $T_2 = 70^0 C$ the density of the water is $\rho = 977.8 \ kg/m^3$

The mass of the water is $\rho V = \rho _1 V_1 = \rho _2 V_2$

Thus; we can say $\rho _1 V_1 = \rho _2 V_2$ ;

⇒ $999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2$

$V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }$

$V_2 = 1022.40 \ mL$

$v_2 = 1022400 \ mm^3$

Thus, the volume of the water after heating to a required temperature of $70^0C$ is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

$A_1 = 1000 \ mm^2$

The area of the heated water is :

$A_2 = A_1 (1 + \Delta t \alpha )^2$

$A_2 = A_1 (1 + (T_2-T_1) \alpha )^2$

$A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2$

$A_2 = 1000.5 \ mm^2$

Finally, the depth of the heated hot water is:

$h_2 = \dfrac{V_2}{A_2}$

$h_2 = \dfrac{1022400}{1000.5}$

$\mathbf{h_2 =1021.9 \ mm}$

Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

4 0

3 years ago

Steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a

Veronika [31]

Answer:

a. 6 seconds

b. 180 feet

Explanation:

Images attached to show working.

a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.

b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.

7 0

3 years ago