A53.6 g mass is attached to a horizontal spring with a spring constant of 8.84 n/m and released from rest with an amplitude of 21.3 cm. what is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless? answer in units of m/s.

The speed of the mass halfway to the equilibrium position is 0.55 m/s.

Explanation:

if K is the kinetic energy and P is the potential energy then the total energy of the system is:

Etot = K + P

we given the amplitude, and the halfway distance is x = 1/2×A them we can get Pmax by:

Pmax = P + K

1/2×k×A^2 = 1/2×k×x^2 + 1/2×m×v^2

      k×A^2 = k×x^2 + m×v^2

     m×v^2 = k×(A^2 - x^2)

          v^2 = k×(A^2 - x^2)/m

                 = (8.84)×((21.3×10^-2)^2 - (1/2(21.3×10^-2)^2))

                 = 0.30079647

              v = 0.55 m/s

Therefore, the speed of the mass halfway to the equilibrium position is 0.55 m/s.

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.