Does a general repair manual have the answers to all repairs?

4. What are these parts commonly called?

patriot [66]

These parts are commonly called carburetor emulsion tubes. These tubes maintain the air-fuel ratio at different speeds.

The carburetor is a device of the combustion engine power supply system that mixes fuel and air in order to facilitate internal combustion.

The carburetor emulsion tubes are tubes that maintain the air-fuel ratio at different velocities.

These tubes (carburetor emulsion tubes) are small brass cylinders where the metering needle slides into them.

Learn more about carburetors here:

brainly.com/question/4237015

7 0

2 years ago

Please answer the questions !

gizmo_the_mogwai [7]

Answer:

120

Explanation:

6 0

3 years ago

The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb>ft. Determine t

irina1246 [14]

Answer:

M = 281.25 lb*ft

Explanation:

Given

Wman = 150 lb

Weight per linear foot of the boat: q = 3 lb/ft

L = 15.00 m

Mmax = ?

Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):

∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0

⇒ q' = (W + q*L) / L

⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft

⇒ q' = 13 lb/ft (+↑)

The free body diagram of the boat is shown in the pic.

Then, we apply the following equation

q(x) = (13 - 3) = 10 (+↑)

V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)

M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft

then

M(7.5) = 5(7.5)² = 281.25 lb*ft

8 0

3 years ago

Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin

Marina CMI [18]

Solution:

Given that :

Volume flow is, $Q_1 = 1000 \ mm^3/s$

So, $Q_2= \frac{1000}{100}=10 \ mm^3/s$

Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$ ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4}$

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$ ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

$=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

7 0

3 years ago

A reversible process and an irreversible process both have the same________ between the same two states. a. Internal energy b. W

Vlad1618 [11]

Answer:

a) Internal energy

Explanation:

As we know that internal energy is a point function so it did not depends on the path ,it depends at the initial and final states of process.All point function property did not depends on the path.Internal energy is a exact function.

Work and heat is a path function so these depend on the path.They have different values for different path between two states.Work and heat are in exact function.

We know that in ir-reversible process entropy will increase so entropy will be different for reversible and ir-reversible processes.

5 0

3 years ago