Answer:
0.695s
Explanation:
From Hooke's law, the restoring force is given has
F = -ky .......1
Where F is the force, y is the spring displacement and k force constant of the spring.
Also recall,
F=mg ............ 2
Where m is the mass of object, g is the acceleration due to gravity.
Equating 1 and 2
Ky = mg
Given that g=9.8m/s2 , y is 3.4cm and g is 8g
K×3.4/100m =8/1000kg × 9.8m/s2
K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034
K= 0.0784÷0.035
K=2.24N/m
Mass ofvthe second object is 25g =0.025kg
Period of oscillation T
T=2π√m/k
T=2×3.142√0.025/2.24
T=6.284√0.0111
T=0.659seconds
Answer:
0.62 m/s² at 68° S of E
Explanation:
Net force north = 12 - 70 = -58 N
Net force east = 33 - 10 = 23 N
Net force = √(-58² + 23²) = 62.3939... N
acceleration = F/m = 62.3939/100 = 0.623939... ≈ 0.62 m/s²
θ = arctan(-58/23) = -68.3691... ≈ 68° S of E
T = 4.25 ms = 4 x 10⁻³ s, the time for rebound
v₁ = 25.5 m/s, the impacting velocty
v₂ = -19.5 m/s, the rebounding velocity (n the opposite directon)
The change in velocity is
v₂ - v₁ = - (25.5+19.5) = -45 m/s
The acceleration is
a = (-45 m/s)/(4 s) = -11.25 m/s²
The negative sign indicates that the final velocity is opposiye to the impact velocty.
Answer: The magnitude of the acceleration is 11.25 m/s²
Answer:
yes
Explanation:
its not a good thing for the rest of your life but you have a PS4
Answer: P = 36.75W
The additional power needed to account for the loss is 36.75W.
Explanation:
Given;
Mass of the runner m= 60 kg
Height of the centre of gravity h= 0.5m
Acceleration due to gravity g= 9.8m/s
The potential energy of the body for each step is;
P.E = mgh
P.E = 60 × 9.8 × 0.5
PE = 294J
Since the average loss per compression on the leg is 10%.
Energy loss = 10% (P.E)
E = 10% of 294J
E = 29.4J
To calculate the runner's additional power
given that time per stride is = 0.8s
Power P = Energy/time
P = E/t
P = 29.4J/0.8s
P = 36.75W
