Answer:
so first wrote down
Explanation:
that will be concluded as the answer
Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°
Hipparcos estimated its distance at roughly 96 parsecs from Earth, or 310 ± 20 light years away.
Answer:
5.4 ms⁻¹
Explanation:
Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.
= length of the meter stick = 1 m
= mass of the meter stick
= angular speed of the meter stick as it hits the floor
= speed of the other end of the stick
we know that, linear speed and angular speed are related as
= height of center of mass of meter stick above the floor = 
= Moment of inertia of the stick about one end
For a stick, momentof inertia about one end has the formula as
Using conservation of energy
Rotational kinetic energy of the stick = gravitational potential energy
