A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.200 m and the period is 3.35 s . a.) What is the acceleration when x = 0.160? b.) What is the speed of the block when x= 0.160 m ?

a = 0.563 m/s²

v = 0.225 m/s

Explanation:

The angular frequency is:

ω = 2π / T

ω = 2π / (3.35 s)

ω = 1.88 rad/s

For a spring mass system:

ω = √(k / m)

ω² = k / m

(a) Sum of forces on the mass:

∑F = ma

kx = ma

a = kx / m

a = xω²

When x = 0.160 m:

a = (0.160 m) (1.88 rad/s)²

a = 0.563 m/s²

(b) Energy is conserved.  Initial elastic energy = final elastic energy + kinetic energy.

EE₀ = EE + KE

½ kA² = ½ kx² + ½ mv²

½ (k/m)A² = ½ (k/m)x² + ½ v²

½ ω²A² = ½ ω²x² + ½ v²

ω²A² = ω²x² + v²

v² = ω² (A² − x²)

v = ω √(A² − x²)

v = (1.88 rad/s) √((0.200 m)² − (0.160 m)²)

v = 0.225 m/s

answer; i believe the correct answer is (40,000 j) good luck

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.