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:78.5 cm^2
Step-by-step explanation:A = pi xr^2
3.14x5x5
-50 much colder now than 75
Answer:
I think there would be 86 blue marbles
x=4(106 - x ) + 6
x= 424 - 4x + 6
5x= 430
x= 86 blue marbles
Tell me if I'm wrong, hope this helps.
g(x) = x^2 + 1(3x - 5)
<em><u>Replace all x's with 4.</u></em>
g(4) = 4^2 + 1(3 * 4 - 5)
g(4) = 16 + 1(12 - 5)
g(4) = 17(7)
g(4) = 119 (This is your answer.)
