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timofeeve [1]
3 years ago
6

During the stage of internal combustion engine operation in which the piston rises and compresses the fuel in the combustion cha

mber,
a. both the intake and exhaust valves are closed.
b. both the intake and exhaust valves are open.
c. the exhaust valve is open and the intake valve is closed.
d. the intake valve is open and the exhaust valve is closed.
Physics
2 answers:
katovenus [111]3 years ago
5 0

The piston rises and compresses the fuel in the cylinder during
the "compression" stroke. 

Both valves have to closed in order to compress the fuel in there. 

If either valve were open, just a crack, then the gases wouldn't get
as compressed as they should ... they'd just get forced out into the
fuel line or the exhaust line through the open valve.

Leokris [45]3 years ago
5 0

your answer is a. both the intake and exhaust valves are closed.

i just took the test

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Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

6 0
3 years ago
Determine the kinetic energy of a 10 Kg roller coaster car that is moving with a speed of 2m/s.
skelet666 [1.2K]

In this question all required information's are already provided. Based on these details the answer to the question can be easily determined. Let us now write down all the information's that are already given.

Mass of the roller coaster = 1000 kg

Velocity of the roller coaster = 20.0 m/s

We know the formula for finding the kinetic energy is

Kinetic energy = 0.5 * mass * (velocity) ^2

                      = 0.5 * 1000 * (20)^2

                      = 0.5 * 1000 * 400

                      = 200000 Joules

So the Kinetic energy of the roller coaster is 200000 joules.

i hope this helps you friend good luck on your quiz or lesson

6 0
2 years ago
A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
MissTica

Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )

\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )

\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )

To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

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Natalka [10]
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for more info , comment !
5 0
3 years ago
A toy rocket fires its engines to launch it straight up from rest. After a short time, the engine turns off and the
madreJ [45]

Answer:

at 10 seconds.. b/c that's when the velocity stops increasing..

Explanation:

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