After a new

A large increase in elevation can cause a carbureted engine to run ________ if not properly adjusted for the altitude. a Rich b

mash [69]

Answer:

B - Poor

Explanation:

As you get higher up, There is less oxygen which causes the engine to create less power.

3 0

3 years ago

A solid circular cylinder of mass m, radius r, and length l is pivoted about a transverse axis

SashulF [63]

To be honest i have no idea

8 0

3 years ago

(Almost) FREE POINTS!!!!! Hey there, Im writting a research paper, and I want to know if you think NASA is a waste of money or n

tatuchka [14]

Answer:

No I don't think NASA is a waste of time. I don't think its a waste of time because NASA is exploring what is beyond Earth and its important to know about Space.

Explanation:

3 0

3 years ago

Read 2 more answers

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$

We can assume the same drag coeeficient $C_{D1}=C_{D2}=C_{D}$ and we got:

$\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D} \frac{\rho}{2}V^2 (wh) V$

$\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})$

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

$R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}$

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

$\frac{l}{h}=\frac{5}{1.7}2.941$

And we can estimate the calue of $C_D = 1.2$ from a figure.

And we can calculate the power difference like this:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

8 0

3 years ago

Find the current in 1 the electrical circuit below

Lesechka [4]

Answer:

c. 25 μA

Explanation:

The dependent current source means that 31 times i1 will flow through the 6kΩ resistor, effectively increasing its voltage drop to 31 times that which would result from i1 alone. In effect, the 6kΩ resistor behaves in the left-side circuit loop as though it were 31×6kΩ = 186kΩ (with no dependent current source).

Then the current i1 is equivalent to that created by a 5+1 = 6V source through a 54kΩ +186kΩ = 240kΩ circuit impedance.

(6V)/(240kΩ) = 25 μA

_____

<em>Additional comment</em>

The voltage across the 6kΩ resistor is (186/240)·6V = 4.65V, and the 25 μA current generates a voltage of 30·(25 μA)(1.8kΩ) = 1.35V across the 1.8kΩ resistor. This means the voltage source at the right side of the diagram needs to be at least 4.65 +1.35 = 6.0V in order to support the calculated voltage drops.

3 0

3 years ago