Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
Answer:
Acceleration of that planet is 30 
.
Given:
initial speed of hammer = 0 
time = 1 s
distance = 15 m
To find:
Acceleration due to gravity = ?
Formula used:
Distance covered by hammer is given by,
s = ut + 
s = distance
u = initial speed of hammer
t = time taken by hammer to reach ground
a = acceleration
Solution:
Distance covered by hammer is given by,
s = ut +
s = distance
u = initial speed of hammer
t = time taken by hammer to reach ground
a = acceleration
u = 0
t = 1 s
s = 15 m
a = g
Thus substituting these value in above equation.
15 = 0 + 
g = 15 × 2
g = 30
Thus, acceleration of that planet is 30 .
The answer is d I took the test
145 Grams!
It asks for the “Total Mass” basically asking to add, If you add 20 to 125, you get 145! Correct me if im wrong
