Let,
Coefficient of viscosity of liquid = eta
Radius of sphere = a
velocity of sperical body = v
Dimensionally it can be proved that
F = k eta*a*v
Experimentally Stoke found that
k = 6*pi
therefore, F = 6*pi*eta*a*v
This is Stoke's law.
the tangential velocity is equal to the angular velocity multiplied by the radius the tangential velocity is equal to the angular velocity multiplied by the radius
Use the formula for centripetal acceleration: velocity squared / radius.
velocity: v = 121 m/s radius: r = 4230m circumference: C = 2 * (pi) * radius = ~26578m time: t = s/v = distance travelled / velocity = circumference / velocity = 220 s
On an irregularly shaped conductor, the surface charge density is greater at locations where the radius of curvature is smallest due to the electric field created by the surface charges. According to Gauss's law, the electric field strength is inversely proportional to the radius of curvature; sharper points or edges (smaller radius) produce a stronger electric field. To maintain electrostatic equilibrium, this results in a higher concentration of charge at these points, leading to greater surface charge density where the curvature is smallest. Thus, the distribution of charge is not uniform, reflecting the geometry of the conductor.
Density = mass/ volume volume= 4/3(pie)(r^3) ***r= radius in meters** so find volume then divide mass by volume and there you go.
I believe the relationship between the radius and velocity is: V = (127 2⁄15 x f x R)½ Where V = velocity, f = lateral load factor, R = Radius I believe the relationship between the radius and velocity is: V = (127 2⁄15 x f x R)½ Where V = velocity, f = lateral load factor, R = Radius
The escape velocity of a particle of mass m is independent of the mass of the particle. It is solely dependent on the mass and radius of the object it is trying to escape from. The escape velocity is given by the formula: (v = \sqrt{\frac{2GM}{r}}), where G is the gravitational constant, M is the mass of the object, and r is the distance from the center of the object to the particle.
An alpha particle is a positively charged particle, so it will experience a force perpendicular to both its velocity and the magnetic field direction. This force causes the alpha particle to move in a circular path due to the magnetic field's influence. The radius of the circle will depend on the velocity of the alpha particle and the strength of the magnetic field.
No, its depends on the planets gravitational pull
The period of the second particle with the same electric charge but ten times as massive will be the same as the first particle in the same field as the period of a particle in a uniform electric field is not dependent on its mass.
Assuming equal velocity. The alpha particle has twice the charge but four times the mass so it would have the wider radius.
Increase in radius affect the increase of the centripetal force on a particle in uniform circular motion. An increase in radius would cause a decrease in the force if velocity remains constant.
Linear velocity is directly proportional to the radius at which the object is moving and the angular velocity of the object. The equation that represents this relationship is v = rω, where v is the linear velocity, r is the radius, and ω is the angular velocity. As the angular velocity increases, the linear velocity also increases, given the same radius.
In circular motion, velocity is directly proportional to the radius and angular velocity (omega). This means that as the radius or angular velocity increases, the velocity of the object in circular motion also increases.
The velocity of a rotating object is directly proportional to its radius. As the radius increases, the velocity also increases to maintain angular momentum. Mathematically, this relationship is described by the equation v = rω, where v is the linear velocity, r is the radius, and ω is the angular velocity.
Angular velocity is inversely proportional to the radius of rotation. This means that as the radius increases, the angular velocity decreases, and vice versa. Mathematically, the relationship can be expressed as ω = v/r, where ω is the angular velocity, v is the linear velocity, and r is the radius.
the velocity will be velocity divided by square root of 2