They are both sites of new cell formation.
Answer:
The circuit is a parallel circuit as the voltage is splitting in equal halves for the light bulbs
Answer:
The horizontal component of the truck's velocity is: 23.70 m/s
The vertical component of the truck's velocity is: 3.13 m/s
Explanation:
You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.
The identities are:
Cosα= ![\frac{CA}{H}](https://tex.z-dn.net/?f=%5Cfrac%7BCA%7D%7BH%7D)
Senα= ![\frac{CO}{H}](https://tex.z-dn.net/?f=%5Cfrac%7BCO%7D%7BH%7D)
Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus
The horizontal component of the truck's velocity is:
Let Vx represent it.
In this case, CA=Vx, H=24 and α=7.5 degrees
Vx=(24)Cos(7.5)
Vx=23.79 m/s
The vertical component of the truck's velocity is:
Let Vy represent it.
In this case, CO=Vy, H=24 and α=7.5 degrees
Vy=(24)Sen(7.5)
Vy=3.13 m/s
Answer:
, assuming that the speed of the electron stays the same.
Explanation:
Let
denote the speed of this electron. Let
denote the electric charge on this electron. Let
denote the mass of this electron.
Since the path of this electron is a circle (not a helix,) this path would be in a plane normal to the magnetic field.
Let
denote the strength of this magnetic field. The size of the magnetic force on this electron would be:
.
Assuming that there is no other force on this electron. The net force on this electron would be
. By Newton's Second Law of motion, the acceleration of this electron would be:
.
On the other hand, since this electron is in a circular motion with a constant speed:
.
Combine the two equations to obtain a relationship between
(radius of the path of the electron) and
(strength of the magnetic field:)
.
Simplify to obtain:
.
In other words, if the speed
of this electron stays the same, the radius
of the path of this electron would be inversely proportional to the strength
of the magnetic field. Doubling the radius of this path would require halving the strength of the magnetic field (to
.)