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Vikentia [17]
3 years ago
8

Most cars today use an ___ combustion engine ... A. internal B. external

Physics
2 answers:
Sphinxa [80]3 years ago
5 0

Answer:

A. internal

Explanation:

Engine is the name given to the equipment responsible for turning energy into motion of the vehicle in which it is installed. In the case of cars and other vehicles, the engine is called “internal combustion” because it transforms the burning of fuels in its interior in motion.

Regardless of engine type, everyone needs the same elements: air, fuel, and an explosion trigger. In Otto engines, this trigger is provided by the spark plug, which unleashes a spark inside the cylinder and generates the explosion. In diesel engines, such an explosion is generated by the compression of gases within the cylinder and the elevated temperature.

Greeley [361]3 years ago
4 0

A. Internal. Most cars use that type of set up because it's more efficient, you can find more about it on this website, https://auto.howstuffworks.com/did-cars-ever-have-external-combustion-engines.htm  

:)

~ Ria

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A car travels 15 kilometers west in 10 minutes. After reaching the destination, the car travels back to the starting point, agai
jeka94

Speed = (distance traveled) / (time to travel the distance).
 
Strange as it may seem, 'velocity' is completely different. 

Velocity doesn't involve the total distance traveled at all. 
Instead, 'velocity' is based on 'displacement' ... the distance
between the start-point and end-point, regardless of the route
taken to get there.  So the displacement in driving once around
any closed path is zero, because you end up where you started. 

Velocity =

           (displacement during some time)
divided by
            (time for the displacement)

AND the direction from the start-point to the end-point.


For the guy who drove 15 km to his destination in 10 min, and then
back to his starting point in 5 min, (assuming he returned by way of
the same 15-km route):

         Speed = (15km + 15km) / (10min + 5min)  =  (30/15) (km/min)

                                                                                 =  2 km/min.

        Velocity = (end location - start position) / (15 min)  =  Zero .

5 0
3 years ago
Determine the accelerations that result when a 45 N net force is applied to 3kg object
svp [43]

Answer:

acceleration is 18

Explanation:

45N/3kg=18

4 0
3 years ago
Read 2 more answers
While filming an intense action sequence for the next James Bond movie, a controlled explosion detonates 1.1 km away from the ac
kolezko [41]

Answer:

The time that will pass between the feeling and hearing the explosion is 2,86 secs

Explanation:

First, let's calculate the time that the wave takes to travel until the actors feel the explosion:

1,1 km*\frac{1.000 mts}{1 km} *\frac{sec}{3.000 mts} = 0,37 secs

Now, the time that pass while the actors hear the sound is:

<em>(Remember that the sound speed in the air is 340 m/s on average)</em>

1,1 km * \frac{1.000 mts}{1 km} * \frac{sec}{340 mts} = 3,23 secs

So, the time between the feeling and hearing is 3,23 - 0,37 = 2,86 secs

4 0
4 years ago
What is the thermal efficiency of a gas power cycle using thermal energy reservoirs at 627°C and 90°C?
vlada-n [284]

Answer:

The value is \eta = 63\%

Explanation:

Generally the thermal efficiency is mathematically represented as

\eta =  1 -  \frac{T_L}{T_H} * 100

substituting  [ 627°C + 273  =  900K ]  for  T_H and    [ 90°C + 273  =  333K ]  for T_L

   So  

         \eta =  1 -  \frac{333}{900} *100

=>     \eta = 63\%

5 0
4 years ago
The de broglie wavelength of an electron with a velocity of 6.00 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-
WINSTONCH [101]

Answer: 1.212(10)^{-10} m

Explanation:

The de Broglie wavelength \lambda is given by the following formula:

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Where:

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p is the momentum of the atom, which is given by:

p=m_{e}v (2)

Where:

m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg is the mass of the electron

v=6(10)^{6}m/s is the velocity of the electron

This means equation (2) can be written as:

p=(9.11(10)^{-31}kg)(6(10)^{6}m/s) (3)

Substituting (3) in (1):

\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)} (4)

Now, we only have to find \lambda:

\lambda=1.2122(10)^{-10} m>>> This is the de Broglie wavelength of the electron

8 0
3 years ago
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