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prohojiy [21]
3 years ago
14

How strong is naruto i need a in depth answer going off his AP and DC i prefer powerscalers since they do this but anyone can an

swer i heard hes Universal AP since he dmg kaguya please HELP!!1!
Physics
1 answer:
marin [14]3 years ago
6 0

Answer:

Naruto is kinda strong

Explanation:

If were just talking about just him I think that he is strong for letting go of his past and moving along with his life. Bu t he also isn't strong if you think about it if Naruto didn't have the nine tails he wouldn't be special so he also is not strong there for his only power really comes from the nine tails.

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Answer:

D

Explanation:

erm i think its 225 ft?

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4 years ago
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Which of the following refers to a force acting toward the center of a circular
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D. Centripetal Force :)

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You drop a basketball (Mass = 0.5 kg) and a medicine ball (mass = 5 kg) from the window of the leaning tower of Pisa. Assuming t
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Two blocks with rough surfaces are stacked on top of a slippery horizontal surface. The top block has a mass of m, and the botto
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Answer:

Please refer to the attached schematic for free-body diagrams of the blocks.

The contact forces N1 = mg, N2 = 3mg.

The friction force fs = mg/2.

The tension force T = 3mg/2.

The distance y = (gt^2)/4

Explanation:

Starting with the third block which is declining a time t. Newton's Second Law implies that

F = (3m)a\\3mg - T = 3ma\\T = 3mg - 3ma

There are two unknowns in one equation: T and a. Therefore, we need to find more equations to solve these unknown variables.

Let's look at free-body diagram 1:

Newton's Second Law:

F = ma\\f_s = ma

That is another equation with one more unknown: fs.

Free-body diagram 2:

F = (2m)a\\T - f_s = 2ma

That is our third equation. We now have three equations with three unknowns. Combining equation 2 and 3 gives:

T - ma = 2ma\\T = 3ma

Plugging this equation into the first equation gives

T = 3mg - 3ma\\3ma = 3mg - 3ma\\3mg = 6ma\\a = g/2

Following this, we can find the other unknowns:

T = 3ma = 3m(g/2) = \frac{3mg}{2}

f_s = ma = mg/2

The contact forces are N1 and N2.

N_1 = mg\\N_2 = 2mg + N1 = 3mg

Finally, using the acceleration and equation of kinematics, we can find the distance the hanging weight drops in a time t.

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6 0
4 years ago
A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 592 N. As the elevator la
gladu [14]

Answer:

<h2>a) 496N</h2><h2>b) 50.56kg</h2><h2>c) 1.90m/s²</h2>

Explanation:

According to newton's secomd law, ∑F = ma

∑F is the summation of the force acting on the body

m is the mass of the body

a is the acceleration

Given the normal force when the elevator starts N1 = 592N

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When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)

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Adding equation 1 and 2 we will have;

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592N + 400N = 2Fg

992N 2Fg

Fg = 992/2

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The weight of the person is 496N

<em>\b) To get the person mass, we will use the relationship Fg = mg</em>

g = 9.81m/s

496 = 9.81m

mass m = 496/9.81

mass = 50.56kg

c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;

N1-N2 = 2ma

592-400 = 2(50.56)a

192 = 101.12a

a = 192/101.12

a = 1.90m/s²

3 0
4 years ago
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