A)♣ the string being pulled,
<span>the angular speed is: w1=w0 +w’*t, hence t=(w1-w0)/w’; </span>
<span>the angular path is: b=w0*t+0.5*w’*t^2, where angular acceleration </span>
<span>w’=T/J, torq T=F*r, and b*r=L; </span>
<span>♦ thus b=w0*(w1-w0)/w’ +0.5*w’*((w1-w0)/w’)^2 = </span>
<span>= (w1-w0)*(w0 +0.5w1 -0.5w0)/w’ =0.5*(w1^2 –w0^2)/w’; </span>
<span>♠ and b=L/r =0.5*(w1^2 –w0^2)/(F*r/J); </span>
<span>2(F*r/J)*L/r =w1^2 –w0^2, hence </span>
<span>w1^2=2F*L/J +w0^2; </span>
<span>b)♣ the power is P=F*(w0*r);
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Answer:
The work done by a magnetic feild on a chraged particle moving in it is zero
Answer:
<h3>C. </h3>
Explanation:
PO NG SAGOT
<h2>#MARK BRAINLITS PO</h2>
TINUTULUNGAN KO PO KAYO
Answer:
Centre of mass of any body is a point where all mass of a body is supposed to be concentrated
it lies in geometrical centre....
Answer:
= 1,386 m / s
Explanation:
Rocket propulsion is a moment process that described by the expression
- v₀ = 
ln (M₀ / Mf)
Where v are the velocities, final, initial and relative and M the masses
The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)
We consider that the rocket starts from rest (v₀ = 0)
At the time of burning half of the fuel the mass ratio is that the current mass is
M = 2.5 Mf
- 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2
= 1,386 m / s
