Efficiency = Work Output / Work Input
92% = Work Output / 100
0.92 = Work Output / 100
Work Output = 0.92 * 100
Work Output = 92 joules.
Answer:
Immediate, potential
Explanation:
In america there are many safety council .in which drivers are trained. According to american safety council in america the drivers are trained in such a way that they can ahead two seconds so that there will not be any immediate hazards and 10 to 12 seconds down the road for potential hazards
So in the blanks there will be immediate and potential
Answer:
Light does not need a medium to travel travel through, but since waves must have a medium to vibrate, sound is not created where no air is present.
Explanation:
Answer:
W = 1418.9 J = 1.418 KJ
Explanation:
In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:
W = F.d
W = Fd Cosθ
where,
W = Work Done = ?
F = Force = 151 N
d = distance covered = 10 m
θ = Angle with horizontal = 20°
Therefore,
W = (151 N)(10 m) Cos 20°
<u>W = 1418.9 J = 1.418 KJ</u>
Wow ! I understand your shock. I shook and vibrated a little
when I looked at this one too.
The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.
"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.
The real question is:
What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?
Acceleration of gravity is
G · M / R²
= (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²
= (6.67 x 10⁻¹¹ · 1.1 x 10³¹ / 4 x 10⁶) (N) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ (kg · m / s²) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ m / s²
That's about 1.87 x 10¹³ times the acceleration of gravity on
Earth's surface.
In other words, if I were standing on the surface of that neutron star,
I would weigh 1.82 x 10¹² tons, give or take.
