1) True 2) True 3)True I hope I helped
<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>
Average speed = (1/2) (beginning speed + ending speed)
= (1/2) ( 13 m/s + 30 m/s )
= (1/2) ( 43 m/s )
= 21.5 m/s
Answer:
the spring compressed is 0.1878 m
Explanation:
Given data
mass = 3 kg
spring constant k = 750 N/m
vertical distance h = 0.45
to find out
How far is the spring compressed
solution
we will apply here law of mass of conservation
i.e
gravitational potential energy loss = gain of eastic potential energy of spring
so we say m×g×h = 1/2× k × e²
so e² = 2×m×g×h / k
so
we put all value here
e² = 2×m×g×h / k
e² = 2×3×9.81×0.45 / 750
e² = 0.0353
e = 0.1878 m
so the spring compressed is 0.1878 m