a force of 100 n applied to an object produces an acceleration of 4.00 m/s2. determine the force needed to give the object an acceleration of 16.0 m/s2
hi trinity, let's do this one!newton's famous second law states, in math terms: f = ma applying to the first case, we have: 100 = m(4.00) => m = 100/4 = 25 kgnow apply the formula to the second case: f = 25 x 16 = 400 newtons
a bob of mass m is suspended by inextensible string of length l from a fixed point. the bob is given a speed of at the bottom point. find the tension in the string when string deflects through an angle 120 degree from vertical. u (initial velocity) = 18 km/hr 
t ( time )= 2.5 sec
v( final velocity) = 0 m/s
from, the first equation of motionv = u +  a×t0 = 5 +  a×2.5a = - 2 m/s2 physics 101 notes: 8/25 (week 2 continued)  force o defined as a push or a pull, and is a vector quantity, meaning that it has direction depending on which way it is being pushed or pulled o it has amount and direction (how much force is being applied and in which direction it is moving)  newton’s first law o first draft: “an object that is free of forces move in a straight line and covers equal distances in equal time” o final: “an object that is free of external forces moves at a constant velocity”  how does a skater start, stop, or turn? o acceleration: “the rate at which an object’s velocity is changing with time o acceleration, s 100 centripetal force:f=mv2rr is equal to l in this question.1. f = (100)(1602)/2500 = 1024 n2. f = (300)(802)/10000 = 192 n3. f = (400)(802)/2500 = 1024 n[readmore]4. f = (200)(1602)/5000 = 1024 n5. f = (800)(402)/10000 = 128 n6. f = (200)(1202)/5000 = 576 n1 = 3 = 4 > 6 > 2 > 5the order of forces acting on the satelites from the largest to the smallest is 1 = 3 = 4 > 6 > 2 > 5 setup the kc expression for the reaction, recall that kc expression will appear as:kc expression for the problem:solving for [n2]:note that ice table is not necessary for this calculation answer is 0.016 m 105 mpa [readmore]typed answera) δg° = -rtlnksp δg° = -(8.314 jmol-k)(298k)ln(4.6×10-2 )= 7628.72 j = 7.63 kjb) ∆g° = nfe°e° = ∆g°-nfe° = 7.63-(2)(96,000)(11000)e° =-0.0397 v answer: `1.00` Δl =l αΔt =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm