Answer:
96.5 km/h
Explanation:
The average velocity of the train is given by:
where
d is the displacement
t is the time taken
For this train, we have:
d = 55 km south (displacement is a vector, so we must also consider the direction)
Substituting into the equation, we find the average velocity:
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
A is correct according to below calculation.
m₁v₀₁+m₂v₀₂=m₁v₁+m₂v₂
((m₁v₀₁+m₂v₀₂)-m₁v₁)/m₂=v₂
v₂=((.5*12-.75*16)+(.5*21.6))/.75
v₂=6.4 m/s
Answer:
Explanation:
Use the equation:
(where T is the period)
(9 comes from the period of a full wavelength)
It’s 180 m/s^2 dude. I think I have you in my class lol.
