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4 years ago

Which stage occurs just before ignition in an internal combustion engine?

2 answers:

6 0

usually, the fuel pump turns on, along with all of the vehicle's electric systems, but on a large diesel engine, like in a semi, or agricultural equipment, the intake, and carburetor are activated first to allow diesel (fuel) into the engine to combustion can be produced once the engine turns over

ddd [48]4 years ago

5 0

I think it is this because compression stroke it needs to be compressed then open up when started.

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Planck's constant, h, is 6.626 x 10-34 js. the speed of light in a vacuum, c, is 3.00 x 108 m/s. calculate the frequency of a gr

Misha Larkins [42]

Frequency= velocity of light/wave length
Fr= 3×10^8/510×10^-9
Frequwency=5.88×10^14 Hz

4 0

3 years ago

A man starts his car from rest and accelerates at 1 m/s2 for 2 s. He then continues at a constant velocity for 10 s until

Bumek [7]

Answer: I showed you all calculation . You did not attach any graph to question .

Explanation:

Lets first find Velocity

Vr=o m/s

Ve=?

a=1m/s²

t=2s

----------

a=(Vr-V)/t

1m/s²=Vr-0m/s/2s

2m/s=Vr

Lets find the time neeeded to stop :

a=1m/s²

Vs=2m/s

Vf=0m/s

a=(Vf-Vs)/t

t*1m/s²=2m/s

t=2 s

4 0

3 years ago

Read 2 more answers

Explain how a refrigerator work to cool down warm object that would otherwise be room temperature

SashulF [63]

The compressor constricts the refrigerant vapor, raising its pressure, and pushes it into the coils on the outside of the refrigerator. 2. When the hot gas in the coils meets the cooler air temperature of the kitchen, it becomes a liquid. ... The refrigerant absorbs the heat inside the fridge, cooling down the air.Sep 22, 2017

7 0

3 years ago

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

never [62]

Answer:

1) Vf = 3.36 m/s

2) v = 6.86 m/s

3) s = 92.3 m

4) a = - 0.23 m/s²

Explanation:

We use first equation of motion in this case:

Vf = Vi + at

where,

Vf = Final velocity = ?

Vi = Initial velocity = 0 m/s

a = acceleration = 1.4 m/s²

t = time = 2.4 s

Therefore,

Vf = 0 m/s + (1.4 m/s²)(2.4 s)

Vf = 3.36 m/s

We again use first equation of motion but with t= 4.9 s now:

Vf = 0 m/s + (1.4 m/s²)(4.9 s)

Vf = 6.86 m/s

Now, this velocity remained constant for net 11 seconds. Hence, the velocity of hare after 8.9 s is:

v = 6.86 m/s

First we use second equation of motion to find distance covered in accelerated motion:

s₁ = Vi t + (0.5)at²

s₁ = (0 m/s)(4.9 s) + (0.5)(1.4 m/s²)(4.9 s)²

s₁ = 16.8 m

Now, we calculate the distance covered in uniform motion:

s₂ = vt

s₂ = (6.86 m/s)(11 s)

s₂ = 75.5 m

Now, total distance covered before slowing down is given as:

s = s₁ + s₂ = 16.8 m + 75.5 m

s = 92.3 m

Using third equation of motion for the decelerated motion:

2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance covered = 102 m

Vf = Final Speed = 0 m/s

Vi = Initial Speed = 6.86 m/s

Therefore,

(2)(a)(102 m) = (0 m/s)² - (6.86 m/s)²

a = - (47.06 m²/s²)/(204 m)

a = - 0.23 m/s²

3 0

4 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

Dafna1 [17]

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$