James observes that the Polaris star in the northern hemisphere does not rise and set in the sky. His teacher tells him that thi
s is because Earth rotates around an axis that points toward Polaris. Based on this information, identify where Polaris could be located on the diagram.
2 answers:
Answer:
Just above the pole (top-most red circle)
Explanation:
Polaris is used to identify North direction. Since, the Earth rotates on its axis which is along North-south, Polaris never seems to rise and set from the Northern hemisphere. This is because Polaris lies above north pole. Thus, in the given diagram, Polaris is above the North pole on the axis represented by top-most red circle.
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Answer
given,
Time period= T = 1.5 s
If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s
v_max=1.00 m/s
we know,
v_ max=A ω
v = A sin (ωt)
-0.50= -1.00 sin (ωt)
sin (ωt) = 0.5
t = 0.125 s
we have time period T=1.5 it is the time to complete one oscillation
means from eq to right,then left,then eq,then left,then from right to eq
time taken for left = t/4 = 0.125/4 = 0.375 s
smallest value of time
=0.375 + 0.125
= 0.50 sec
Answer:
:Final block velocity (toward the left)
:Final mass velocity (to the right)
Explanation:
To solve this problem we apply the theory of shocks:
In an elastic shock the kinetic energy and the amount of linear movement or momentum are conserved.
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.
Principle of conservation of the momentum:
m1vi1+m2vi2=m1vf1+m2vf2 Equation 1
Formula to calculate the coefficient of elastic restitution (e):
Equation 2
m1: Block mass
m2: mass of the body that collides with the block
Vi1,vf1: initial, final velocity of the block
Vi2,vf2: initial, final velocity of the body that collides with the block
Of the problem data we know that:
m1=m , m2 = 2m, vi1=v and vi2=0, then, we replace this information in equation (1) :
mv+0=mvf1+2mvf2 we divide by m:
v=vf1+2vf2 Equation (3)
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1,then , we we replace this information in equation (2)
v=vf2-vf1 Equation (4)
vf2=v+vf1 Equation (5)
We replace the equation 5 in the equation (3)
v=vf1+2(v+vf1)
v=vf1+2v+2vf1
-3vf1=v
We replace Vf1=(-1/3)v in the equation (5):
Answer;
:Final block velocity (toward the left)
:Final mass velocity (to the right)