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:
Mass is the same, weight is less.
Explanation:
We know the following equation
W = mg
where m is the mass of the astronaut which is constant and W is the weight. As g (the acceleration due to gravity) is a variable which it is directly proportional to the W, if g is lower on the Moon than on Earth, then W is lower on the Moon than on Earth.
