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3 years ago

Whats goku uchihas quirk

Physics

2 answers:

Sladkaya [172]3 years ago

7 0

Answer:

i think its his rasengan which he can strech any body part

Explanation:

trasher [3.6K]3 years ago

5 0

Answer:All for sage mode

Explanation:

big brain

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A yellow train of mass 100 kg is moving at 8 m/s toward an orange train of mass 200 kg traveling in the opposite direction on th

Serhud [2]

The initial momentum of the yellow and the orange train is 1000kgm/s.

Momentum is the product of the mass and velocity of any object.

Momentum is denoted by P.

Momentum P = mv , where m = mass and v = velocity.

<h3>Given:</h3>

Mass of the orange train = 200kg

Velocity of the orange train = 1m/s

So, the momentum of the orange train will be,

∴ P = mv

P = 200 x 1

P = 200 kgm/s

∴ The initial momentum of the orange train is 200kgm/s.

Mass of the yellow train = 100kg

Velocity of the yellow train = 8m/s

So, the momentum of the yellow train will be,

∴ P = mv

P = 100 x 8

P = 800 kgm/s

∴ The initial momentum of the yellow train is 800kgm/s.

Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.

Learn more about momentum here:

brainly.com/question/25849204

#SPJ1

5 0

2 years ago

A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person

vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 $\frac{m}{s}$

t=0.475s V=1.95 $\frac{m}{s}$

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

$sin(\alpha) =\frac{op}{h}$

$op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m$

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

$T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz$

c).

$T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz$

The period is the inverse of the time of the motion so, the T1 is faster that the T because

$t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s$

d).

The speed is the relation between the distance with time so:

$Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}$

3 0

2 years ago

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t

dybincka [34]

Answer:

A) t = 7.0 s

B) x = 25 m

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1 m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0

3 years ago

You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While

Fynjy0 [20]

Answer:

Explanation:

The question here is that if sneezy hands from a similar rope while delivering presents at the earth's equator, what will be the tension in the rope be. Here is the solution:The tension on the rope when it is at pole, T= 455 NTo find, the tension, t= mgTo solve for mass, m= t/g. Substituting this we have, m=455/9.8. m=46.43 kgAssume that the downwards acceleration is, a= -46.43 m/s^2.T = mg + maT = (46.43 kg) ( 9.8 m/s^2) - (46.43 kg) (-46.43 m/s^2)T = 455.01 kg-m/s^2 - -2155.74 kg-m/s^2T = 2610.75 kg-m/s^2 = 2610.75 N

3 0

2 years ago

Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by

OlgaM077 [116]

Explanation:

Area of ring $\ 2{\pi} a d a$

Charge of on ring $d q=-(\ 2{\pi} a d a)$

Charge on disk

$Q=-\left(\pi R^{2}\right)$

$\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}$

Note: Refer the image attached

8 0

3 years ago