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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!

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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A small airplane has to reach a speed if 27.8 m/s to takeoff. It can accelerate at 2.00 m/s^2. What is the minimal length of run

pickupchik [31]

Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s

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3 years ago

How large must the coefficient of static friction be between the tires and road, if a car rounds a level curve of radius 85 m at

tatuchka [14]

Answer:

0.66

Explanation:

By using the formula

u = v^2 / r g

Where u is coefficent of friction

u = 23.5 × 23.5 / (85 × 9.8)

u = 0.66

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3 years ago

A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is

Crazy boy [7]

Answer:

$v=1.93m/s$

Explanation:

From the concept of fluids mechanics we know that if a tank has a hole at the bottom, the equation that we need to use is:

$v=\sqrt{2gh}$

Since we know gravity and its hight

$v=\sqrt{2*(9.81m/s^{2})(0.19m) }=1.93m/s$

5 0

3 years ago

Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul

Yuki888 [10]

Answer:

$63.4^{\circ}$

Explanation:

When unpolarized light passes through the first polarizer, the intensity of the light is reduced by a factor 1/2, so

$I_1 = \frac{1}{2}I_0$ (1)

where I_0 is the intensity of the initial unpolarized light, while I_1 is the intensity of the polarized light coming out from the first filter. Light that comes out from the first polarizer is also polarized, in the same direction as the axis of the first polarizer.

When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle $\theta$ with respect to the first one, the intensity of the light coming out is

$I_2 = I_1 cos^2 \theta$ (2)

If we combine (1) and (2) together,

$I_2 = \frac{1}{2}I_0 cos^2 \theta$ (3)

We want the final intensity to be 1/10 the initial intensity, so

$I_2 = \frac{1}{10}I_0$

So we can rewrite (3) as

$\frac{1}{10}I_0 = \frac{1}{2}I_0 cos^2 \theta$

From which we find

$cos^2 \theta = \frac{1}{5}$

$cos \theta = \frac{1}{\sqrt{5}}$

$\theta=cos^{-1}(\frac{1}{\sqrt{5}})=63.4^{\circ}$

6 0

3 years ago

A cat is moving at 18 m/s when it accelerates for 2 seconds. What is its acceleration?

vova2212 [387]

Answer: acceleration: $4 m/s^{2}$

velocity: $26 m/s$

Explanation:

The complete question is written bellow:

<em>A cat is moving at 18 m/s when it accelerates at </em> $4 m/s^{2}$ <em> for 2 seconds. What is his new velocity? </em>

<em />

In this situation the following equation will be useful:

$V=V_{o}+at$

Where:

$V$ is the cat’s final velocity (new velocity)

$V_{o}=18 m/s$ is the cat’s initial velocity

$a=4 m/s^{2}$ is the cat's acceleration

$t=2 s$ is the time

Solving the equation:

$V=18 m/s+(4 m/s^{2})(2 s)$

$V=26 m/s$ This is the cat's new velocity

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3 years ago