An object attached to an ideal massless spring is pulled across a frictionless surface. If the spring constant is 45 N/m and the spring is stretched by 0.88 m when the object is accelerating at 0.8 m/s2, what is the mass of the object

Explanation:

Using Hooke's law,

F = -k × delta x

Where,

F = force

k = spring constant

= 45 N/m

Delta x = spring displacement

= 0.88 m

F = 45 × 0.88

= 39.6 N

a = 0.8 m/s^2

Force, F = Mass, M × acceleration, a

M = 39.6/0.8

= 49.5 kg

answer: strength i guess

explanation:

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg