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:
9.081
Step-by-step explanation:
Chow,...!
Okay. On the part that's not the whole number, see how the top number is BIGGER than the bottom? That means that it's not simplified. In order to make it simplified, you have to take a whole out of the pat that's not a whole number. Since the denominator is 12, the whole is 12/12. So, subtract it from the 13/12.
13/12-12/12=1/12
Now, since you found a whole inside of the fraction you have to add 1 to the whole number.
5+1=6
So, the answer is 6 1/12.
Thanks for the points!
Tan80 = QR / RS
QR = 5.67 (2)
QR = 11.34
answer
<span>11.34cm</span>
Answer:
2x + 10
= 2(15) + 10
= 30 + 10
= 40
It will take 40 minutes to make and pack an order for 15 parts.
Step-by-step explanation:
