El bloque de 10 kg que se muestra en la figura está en contacto con el resorte que tiene una longitud natural de 25 cm. Cuando se suelta el bloque, éste comprime al resorte quedando estático a 15 cm del piso. Calcule el cambio de energía potencial elástica del resorte en Joules.

9.81 J

Explanation:

The 10 kg block shown in the figure is in contact with the spring that has a natural length of 25 cm. When the block is released, it compresses the spring, remaining static 15 cm from the floor. Calculate the change in elastic potential energy of the spring in Joules

The given parameters are;

The mass of the block in contact with the spring = 10 kg

The natural length of the spring = 25 cm

The final height of the block after being released = 15 cm

Therefore;

The potential energy of the block = m·g·h

The initial potential energy of the block = 10 kg × 9.81 m/s² × 0.25 m = 24.252 J

The final potential energy of the block = 10 kg × 9.81 m/s² × 0.15 m = 14.715 J

By energy conservation principle

The change in potential energy of the block = The change in elastic potential energy of the spring = (1/2)·k·x² = 24.252 J - 14.715 J = 9.81 J

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s