All categories

3 years ago

During the power stroke energy released by combustion generates the force that:_______.

Physics

1 answer:

Inessa05 [86]3 years ago

8 0

Explanation:

An IC engines means an Internal Combustion engine. Here, the combustion takes place in side a chamber or a cylinder internally, hence it is known as internal combustion engine.

Combustion is the process where oxygen reacts with the fuel and it burns to produce a large amount of force and heat. In each revolution of the crank, the engine creates a power stroke when the piston goes from the bottom dead center to the top dead center.

During each power stroke, the energy released forces the piston down and the crankshaft is rotated.

You might be interested in

LiRa [457]

The horizontal speed of the object 1.0 seconds later is 1) 5.0 m/s.

Explanation:

The motion of an object thrown horizontally off a cliff is a projectile motion, which follows a parabolic path that consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

This means that the horizontal speed of an object in projectile motion does not change, and remains constant during the whole motion.

Since in this case the object has been launched with a horizontal speed of

v = 5.0 m/s

this means that this speed will remain constant during the motion, so its horizontal speed 1.0 s later is also 5.0 m/s.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

Two astronauts of mass 100 kg are 2 m apart in outer space. What is the

fredd [130]

The force of gravity between the astronauts is $1.67\cdot 10^{-7}N$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have two astronauts, whose masses are:

$m_1 = 100 kg\\m_2 = 100 kg$

While the separation between the astronauts is

r = 2 m

Substituting into the equation, we can find the gravitational force between the two astronauts:

$F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

A crowbar having the length of 1.75m is used to balance a load between the fulcrum and the load is 0.5m calculate MA

aliina [53]

Answer:

1.70

Explanation:

substract 1.75-0.5

3 0

2 years ago

A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

ankoles [38]

Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J

4 0

3 years ago

At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh

Afina-wow [57]

Answer:

$W = 1.22 \times 10^9 J$

Explanation:

Initial potential energy of the given spacecraft is given as

$U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}$

so we have

$U = - \frac{Gm}{r}(M_e + M_m)$

so we have

$M_e = 5.98 \times 10^{24} kg$

$M_m = 7.35 \times 10^{22} kg$

$m = 1160 kg$

$r = 3.84 \times 10^8 m$

$U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})$

$U = -1.22 \times 10^9 J$

now total work done to move it to infinite is given

W = 0 - U

$W = 1.22 \times 10^9 J$

6 0

2 years ago