Ah ha ! Very interesting question.
Thought-provoking, even.
You have something that weighs 1 Newton, and you want to know
the situation in which the object would have the greatest mass.
Weight = (mass) x (local gravity)
Mass = (weight) / (local gravity)
Mass = (1 Newton) / (local gravity)
"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest. This is the
clue that gives it away.
If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:
"Fine ! Great ! Golly gee, that's sure generous of you.
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
there. And if you don't mind, be quick about it."
The local acceleration of gravity on Pluto is 0.62 m/s² ,
but on Earth, it's 9.81 m/s.
So if he weighs 1 Newton of gold for you on Pluto, its mass will be
1.613 kilograms, and it'll weigh 15.82 Newtons here on Earth.
That's almost 3.6 pounds of gold, worth over $57,000 !
It would be even better if you could convince him to weigh it on
Halley's Comet, or on any asteroid. Wherever he's willing to go
that has the smallest gravity. That's the place where the largest
mass weighs 1 Newton.
Answer: B. 44.64 g
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Mass of reactants = mass of iron + mass of oxygen = mass of iron + 34.7 g
Mass of product = mass of iron oxide = 79.34 g
As Mass of reactants = Mass of product
mass of iron + 34.7 g = 79.34 g
mass of iron = 44.64 g
Thus 44.64 g of iron was used in the reaction
Answer:
the tension in the string an instant before it broke = 34 N
Explanation:
Given that :
mass of the ball m = 300 g = 0.300 kg
length of the string r = 70 cm = 0.7 m
At highest point, law of conservation of energy can be expressed as :


The tension in the string is:

Thus, the tension in the string an instant before it broke = 34 N
Acceleration = (Vf - Vi)/t
Since Vf= 60m/s
Vi= 15m/s
T= 15s
=> a= (60m/s - 15m/s)/15s
= 3
So the acceleration is 3m/s^2
Answer:
i think the answer is 12 ohms
plz mark me as brainliest :)