Answer:
Your answers would be
4. A. sperm and testosterone.
7. C. prostate, penis, Testes
uterus, vagina, fallopian tubes
10. B. Protein
11. A. carbohydrate
12. B. amino acids (I'm not positive on this i haven't taken bio in years
27. D. Respiratory system
Explanation:
yeah
Answer:
earth
sorry I need points but I don’t know the answer again I’m really sorry
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:
I_weight = M L²
this value is much larger and with it it is easier to restore balance.I
Explanation:
When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by
v = w r
For man to maintain equilibrium needs the total moment to be zero
∑τ = I α
S τ = 0
The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.
Therefore the moment of the masses and the open is the one that must be zero.
If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope
I = ⅓ m L² / 4
As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.
If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is
I_weight = M L²
this value is much larger and with it it is easier to restore balance.