If v (velocity) is constant, then it's straight (linear). If v is changing due to acceleration, then it is a curve (non-linear)
V=wr Where : V: Linear velocity w: Angular velocity r: Radius v = 600 m/m r=0.75m
linear - designated by v cyclic - designated by w
Time dilation, which can be derived from the Lorentz transformations is t'=t/sqrt(1-v^2/c^2) where t is the time interval in the rest frame, and t' is the interval in the lab frame. This relationship is neither linear or exponential in v.
Momentum. The formula for kinetic energy is: KE = .5 * m *v^2 The formula for momentum is: p = m * v If an object has kinetic energy, then both mass and velocity are non-zero, which implies that the momentum is also non-zero.
It appears from the question that the balls stick together after the collision. Linear momentum is conserved. The linear momentum is the total of the product of mass and velocity for each of the balls. The linear momentum before is (1.4 x 3) + (0 x 2) = 4.2 kgms-1. The linear momentum after is v x (3 + 2) = 4.2kgms-1, since we know it is conserved. Hence, v = 4.2 / 5 = 0.84ms-1, in the same direction of travel as the 3kg ball was originally moving.
Zener diode is not a linear device... it is non-linear one. Since linear devices are those devices which have linear characteristics(V-I CHAR.), or follows the Ohm's law i.e. voltage is directly proportional to current. but in case of Zener diode ohm's law fails down. the V-I char. in both forward biased & reverse biased condition is non-linear. So. Zener diode is non-linear device
bent (v-shaped or non-linear)
S = v*t s = displacement v = velocity t = time
V. N. Faddeeva has written: 'Computational methods of linear algebra.'
We know the equation for linear momentum is P = m*v, where P is linear momentum, m is mass and v is velocity. So, we can find the velocity with the other two like this: v = P/m v = 600/50 = 12 m/s.
V=wr Where : V: Linear velocity w: Angular velocity r: Radius v = 600 m/m r=0.75m
Linear algebra deals with mathematical transformations that are linear. By definition they must preserve scalar multiplication and additivity. T(u+v)= T(u) + T(v) T(R*u)=r*T(u) Where "r" is a scalar For example. T(x)=m*x where m is a scalar is a linear transform. Because T(u+v)=m(u+v) = mu + mv = T(u) + T(v) T(r*u)=m(r*u)=r*mu=r*T(u) A consequence of this is that the transformation must pass through the origin. T(x)=mx+b is not linear because it doesn't pass through the origin. Notice at x=0, the transformation is equal to "b", when it should be 0 in order to pass through the origin. This can also be seen by studying the additivity of the transformation. T(u+v)=m(u+v)+b = mu + mv +b which cannot be rearranged as T(u) + T(v) since we are missing a "b". If it was mu + mv + b + b it would work because it could be written as (mu+b) + (mv+b) which is T(u)+T(v). But it's not, so we are out of luck.
N. V. Efimov has written: 'Differentialgeometrie' 'A brief course in analytic geometry' -- subject(s): Analytic Geometry 'Linear algebra and multidimensional geometry' -- subject(s): Analytic Geometry, Linear Algebras
Inductor is a nonlinear device. since v=L di/dt.
linear - designated by v cyclic - designated by w
This is because the resistance is measured by applying a fixed voltage to the resistor and measuring the current. Since I = V/R, the current/resistance relation is non-linear.
Time dilation, which can be derived from the Lorentz transformations is t'=t/sqrt(1-v^2/c^2) where t is the time interval in the rest frame, and t' is the interval in the lab frame. This relationship is neither linear or exponential in v.