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>
There correct answer is B!
<span>The expression of the square root of 19x must be simplified when x is equal to 28. This is because possible factors of 28 can be seen to be 4 and 7, and 4 is a perfect square. This means it can be pulled outside of the square root when evaluated. The other options include only prime factors that could not be pulled out. (3,5), (3,7), (1,41)
28 simplifies as such:
Sqrt(19*28) = Sqrt(19*4*7) = 2*Sqrt(19*7) = 2*Sqrt(133).</span>
Answer:
Step-by-step explanation:
a = 0
So that AOB makes a right triangle with sides 3-4-5
Answer:
g = -3
Step-by-step explanation:
