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Gemiola [76]
4 years ago
6

The ________ is the difference between the volume in the combustion chamber above the piston when the piston is at bottom dead c

enter (BDC) and the volume of the combustion chamber above the piston when the piston is at top dead center (TDC).
Physics
1 answer:
Vitek1552 [10]4 years ago
3 0

Answer: Compression ratio

Explanation:

Let's begin by explaining that in a combustion chamber there are pistons that move up and down within cylinders. If we take as an example one cylinder, we will have the following:

-The bottom dead center (BDC) is when the volume in the combustion chamber is the greatest, because the piston is exactly at the bottom of the cylinder.

- The top dead center (TDC) is  when the volume in the combustion chamber is the lowest, because the piston is at the top (the highest point) inside the cylinder.

Now, the ratio between these two volumes is the Compression ratio.

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Cual es el propósito del ser desde su creación
Ivanshal [37]

Answer:

vivir y servir a Dios

Explanation:

8 0
3 years ago
A marble accelerates from rest at a constant rate .it travels 36 m in 12.0 sec what’s is it’s final velocity and what was the ac
ddd [48]

Answer:

a = 0.50 m/s²

v = 6.0 m/s

Explanation:

Step 1: Given data

  • Initial velocity (u): 0 m/s (rest)
  • Displacement of the marble (s): 36 m
  • Time elapsed (t): 12.0 s
  • Final velocity (v): ?
  • Acceleration (a): ?

Step 2: Calculate the acceleration

We will use the following suvat equation.

s = u × t + 1/2 × a × t²

s = 0 × 12.0 s + 1/2 × a × (12.0 s)²

36 m = a × 72.0 s²

a = 0.50 m/s²

Step 3: Calculate the final velocity

We will use the following suvat equation.

v = u + a × t

v = 0 + 0.50 m/s² × 12.0 s

v = 6.0 m/s

4 0
4 years ago
A plane is flying 2400 miles from a to
geniusboy [140]
From a to b speed is 600+40 = 640
from b to a speed is 600-40 = 560

let t be the number of hours of flight. This would mean it would have traveled a distance of 640 miles and the distance yet to travel is 2400-640t
Time left will be (2400-640t)/640. But if they were to return to a it would fly 640t miles at 560mph which will take (640t/560) hrs

(2400-640t) / 640 = 640t / 560
560(2400 - 640t) = 640t x 640
t = 1.75hrs
5 0
4 years ago
I’m still lost on this please help me on this I know it’s not D I got it wrong
RSB [31]

Before you even look at the questions, look over the graph, so you know what kind of information is there.

The x-axis is "time".  OK.  You know that as the graph moves from left to right, it shows what's happening as time goes on.

The y-axis is "speed" of something.  OK.  When the graph is high, the thing is moving fast.  When the graph is low, the thing is moving slow.  When the graph slopes up, the thing is gaining speed.  When the graph slopes down, the thing is slowing down.  When the graph is flat, the speed isn't changing, so the thing is moving at a constant speed.

NOW you can look at the questions.

OMG !  It's only ONE question:  What's happening from 'c' to 'd' ?  Well I don't know.  Perhaps we can figure it out if we LOOK AT THE GRAPH !

-- Between c and d, the graph is flat.  The speed is not changing.  It's the same speed at d as it was back at c .

What speed is it ?

-- Look back at the y-axis.  The speed at the height of c and d is 'zero' .

-- The 2nd and 4th choices are both correct.  From c to d, <em>the speed is constant</em>.  The constant speed is zero.  <em>The car is not moving</em>.

5 0
3 years ago
There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co
mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

F_{net}=0.208 N

3 0
3 years ago
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