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3 years ago

The main difference between the internal combustion engine and the diesel engine is that the diesel engine

1 answer:

4 0

The diesel engine has no ignition system. (C)
It depends on the compression of the fuel/air mixture to ignite the mixture.

This may be a big part of the reason that Diesel engines always
sound to me as if they are falling apart.

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Please give an answer that is coherent

Semmy [17]

Answer:

480

Explanation:

resistance equals to potential difference divide by electric current

120÷0.25

=480

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2 years ago

Which statement explains the similarity between momentum and kinetic energy of an object?

Anon25 [30]

Answer:B.

Both increase as the mass and velocity increase.

3 0

2 years ago

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A soccer ball is kicked horizontally and rolled off of a 30 meter high cliff. The soccer ball was kicked with a velocity of 17m/

Montano1993 [528]

Answer:

i am doing this rn

Explanation:

COMMENTS

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2 years ago

Pitch marks are the invitations left by the ball when they fall on the green

Kipish [7]

You can mark, lift and clean a ball on the green, but it's a violation to do so when another ball is in motion, as your ball might influence the outcome of that stroke. You can also mark and clean your ball in some instances when it's off the green: cleaning it, for example, just to the point where you can identify it.

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3 years ago

A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa

Darya [45]

Answer:

x=0.53 $x10^{-3} m$

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1. $x10x^{-6} \frac{C}{m^{2}}$

ε=8.85 $x10^{-12} \frac{C^{2}}{n*m^{2}}$

$E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}$

Force of charge is

$F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N$

$F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s$

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

$x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2} \\x_{f}=0.53x^{-3}m$

5 0

2 years ago