Answer:
5/13, 12/5, 12/13
Step-by-step explanation:
cos = adj/hyp
tan = opp/adj
sin = opp/hyp
hyp = 13
adj = 5
opp = 12
Plug in.
cos = 5/13
tan = 12/5
sin = 12/13
The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus,
1.22 a = 12.0 - T (eqn 1)
and for the 20.0 N block:
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction)
2.04 a = T - 6.5 (eqn 2)
[eqn 1] + [eqn 2] → 3.26 a = 5.5
a = 1.69 m/s²
Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5
T = 9.95 N
Now want the resultant force acting on the 20.0 N block:
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N
<span>Units have to be consistent ... so have to convert 75.0 cm to m: </span>
75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m
<span>work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J</span>
Answer:
The answer is 166/21, or 7 19/21
Step-by-step explanation:
First, you would set up the equation, which is:
13 1/7 - 5 5/21
Next, you would turn them into an improper fraction, which is:
92/7 - 110/21
Then, you would multiply 92/7 by 3/3 because that is the least common multiple for 21 and 7 (which are the denominators). When you are done multiplying, the equation would be:
276/21 - 110/21
Afterwards, you would subtract the numerators and leave the denominators alone, which gives you the answer 166/21. As a mixed number, this would equal 7 19/21.
Answer:
87.6
Step-by-step explanation:
