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garri49 [273]
3 years ago
13

An engine with an electronic fuel injection system has high fuel pressure at idle. This could be caused by a:

Physics
1 answer:
olga_2 [115]3 years ago
5 0

Answer:

An engine with an electronic fuel injection system has high fuel pressure at idle because of  high manifold vacuum. (option D)

Explanation:

Electronic fuel injection (EFI) system replaced carburetors back in the mid-1980s as the preferred method of supplying air and fuel to engines. The basic difference is that a carburetor uses intake vacuum and a pressure drop in the venturi, to siphon fuel from the carburetor fuel bowl into the engine. Whereas fuel injection system uses pressure to spray fuel directly into the engine.

However, under light load or at idle, a relatively high vacuum exists in the intake manifold. This means less fuel pressure is needed to spray a given volume of fuel through the injector. Under heavy load, engine vacuum drops to near zero.

Therefore, An engine with an electronic fuel injection system has high fuel pressure at idle because of  high manifold vacuum.

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An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The part
Varvara68 [4.7K]

Answer:

Explanation:

charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg

Radius, r = 4.5 cm = 0.045 m

Magnetic field, B = 1.20 T

(a) Let the speed is v.

v=\frac{Bqr}{m}

v=\frac{1.20\times 3.2\times 10^{-19}\times 0.045}{6.644\times 10^{-27}}

v = 2.6 x 10^6 m/s

(b) Let T be the period of revolution

T=\frac{2\pi r}{v}

T=\frac{2\times 3.14\times 0.045}{2.6\times 10^{6}}

T = 1.09 x 10^-7 s

(c) The formula for the kinetic energy is

K=\frac{B^{2}\times q^{2}\times r^{2}}{2m}

K=\frac{\left ( 1.20\times 3.2 \times 10^{-19}\times 0.045 \right )^{2}}{2\times 6.644\times 10^{-27}}

K = 2.25 x 10^-14 J

(d) Let the potential difference is V.

K = qV

V = \frac{K}{q}

V= \frac{2.25\times 10^-14}{3.2\times 10^{-19}}

V = 70312.5 V

5 0
3 years ago
The peak of intensity for our Sun on a black body curve would be in which of the following wavelength ranges?0.01 – 10nm
saveliy_v [14]

The answer is,

560- 600nm


5 0
3 years ago
Read 2 more answers
A runner maintains a constant velocity of +3.0 m/s. Which of the following position-time graphs shows the runnerÍs location from
Elden [556K]
<span>3 meters a second.
Every 5 seconds, that runner should run 5*3=15 meters
(distance = average velocity times time).
 That means that your slope must be 15meters / 5 seconds. </span>
5 0
3 years ago
a cyclist applies a constant forward force of 20 N to maintain a velocity 2.5 m/s. how much power does the cyclist deliver?
forsale [732]

The power exerted by the cyclist is determined as 50 W.

<h3>Average power exerted by the cyclist</h3>

The power exerted by the cyclist is calculated as follows;

P = FV

where;

  • F is the applied force
  • V is velocity

P = 20 x 2.5

P = 50 W

Thus, the power exerted by the cyclist is determined as 50 W.

Learn more about power here: brainly.com/question/25263760

#SPJ1

4 0
2 years ago
A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

W=K_f -K_i

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving,

K_f = K_i + W=66,120+(-36,733)=29387 J

The final kinetic energy of the car can be written as

K_f = \frac{1}{2}mv^2

where

m = 661 kg is its mass

v is its final speed

Solving for v,

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

4 0
3 years ago
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