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3 years ago

In the four-stroke internal combustion engine, fuel is ignited during the ____ stroke.

2 answers:

7 0

The four strokes in order are the intake stroke, the compression stroke, the power stroke, and the exhaust stroke. Fuel is ignited during the power stroke.

Leona [35]3 years ago

3 0

The power stroke, I believe.
Sorry if this does not help.

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When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coeffic

Nuetrik [128]

Answer:

$8.1\cdot 10^{-4} C^{-1}$

Explanation:

The volumetric expansion of the liquid is given by

$\Delta V=\alpha V_0 \Delta T$

where

$\alpha$ is the coefficient of volume expansion

$V_0$ is the initial volume

$\Delta T$ is the change in temperature

For the liquid in this problem,

$V_0 = 2.35 m^3\\\Delta T=48.5^{\circ}C\\\Delta V=0.0920 m^3$

So we can solve the equation to find $\alpha$ :

$\alpha=\frac{\Delta V}{V_0 \Delta T}=\frac{(0.0920 m^3)}{(2.35 m^3)(48.5^{\circ}C)}=8.1\cdot 10^{-4} C^{-1}$

7 0

4 years ago

An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts

Nimfa-mama [501]

Explanation:

It is given that,

Speed of the electron in horizontal region, $v=1.9\times 10^7\ m/s$

Vertical force, $F_y=4.9\times 10^{-16}\ N$

Vertical acceleration, $a_y=\dfrac{F_y}{m}$

$a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}$

$a_y=5.37\times 10^{14}\ m/s^2$ ..........(1)

Let t is the time taken by the electron, such that,

$t=\dfrac{x}{v_x}$

$t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}$

$t=1.26\times 10^{-9}\ s$ ...........(2)

Let $d_y$ is the vertical distance deflected during this time. It can be calculated using second equation of motion:

$d_y=ut+\dfrac{1}{2}a_yt^2$

u = 0

$d_y=\dfrac{1}{2}\times 5.37\times 10^{14}\ m/s^2\times (1.26\times 10^{-9}\ s)^2$

$d_y=0.000426\ m$

$d_y=0.426\ mm$

So, the vertical distance the electron is deflected during the time is 0.426 mm. Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

The chart shows rate of decay. A 3 column table with 7 rows. The first column is Half-lives elapsed, with entries 0, 1, 2, 3, 4,

Romashka [77]

Answer:

Explanation:

8 0

4 years ago

Read 2 more answers

A 2.40 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N

klemol [59]

Answer: Hello! Here your answer......

Force = 10.244 Newtons

b) No of oscillations = 0.88

Explanation:

Since the block executes SHM we can write it's position as function of time as

ω is the natural frequency of the system

A is the amplitude of the system

Thus accleration of the block

Thus using the given values at t= 3.50 sec we can calculate the acceleration as

thus force can be calculated using newtons second law as

Now no of oscillations can be obtained as

no of oscillations in 3.50 seconds = 3.50/3.976 = 0.88

Hope this helps! (Brainly)♥

3 0

3 years ago

Which of the following statements is accurate?

Sladkaya [172]

There are no accurate statements at all on that list.

5 0

3 years ago