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san4es73 [151]
3 years ago
9

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 project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow fo
Ksju [112]

Answer:

A. The project's energy costs will decrease

Explanation:

Since the project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow for on-site renewable energy.

Hence, the project's energy costs will decrease very well because it's implementing both of these strategies;

- Area with demand-response program.

- On-site renewable energy.

5 0
3 years ago
Which of these are a covalent compound?
nikdorinn [45]
C is a non-metal and so is O. So the answer is CO
8 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
Which conclusion, based on measurements of the ball in the lab in this lesson, is correct? a. The density of the ball was 1.48 g
ser-zykov [4K]

Answer:

u dummy

Explanation:

7 0
2 years ago
honor physics The velocity of a 200 kg object is changed from 5 m/s to 25 m/s in 50 seconds by an applied constant force. a. Wha
Ksivusya [100]

Answer:

(a) 4000 kgm/s.

(b) 80 N

Explanation:

(a) Change in momentum: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is Ns or kgm/s

Mathematically, Change in momentum is expressed as

ΔM = mΔv ..................................... Equation 1

Where ΔM = change in momentum, m = mass of the object, Δv = change in velocity = v₂ - v₁

Given: m = 200 kg, Δv = v₂ - v₁ = 25-5 = 20 m/s.

Substituting into equation 1

ΔM = 200(20)

ΔM = 4000 kgm/s.

Hence the change in momentum = 4000 kgm/s

(b)

Force: This can be defined as the ratio of the change in momentum of a body to the time required for the change.

F = ΔM/t.............................. Equation 2

Where F = force, ΔM = change in momentum, t = time.

Given: ΔM  = 4000 kgm/s, t = 50 second.

Substituting into equation 2

F = 4000/50

F = 80 N.

Hence the force  = 80 N

7 0
3 years ago
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