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:
or
Step-by-step explanation:
we have
This is a vertical parabola open upward (because the leading coefficient is positive)
The vertex is a minimum
Convert to vertex form
Factor the leading coefficient
Complete the square. Remember to balance the equation by adding the same constants to each side.
Rewrite as perfect squares
----> equation in vertex form
or
The vertex is the point (0.25,10.5)
Step-by-step explanation:
g(t) = (t + 1)² − 20.25
0 = (t + 1)² − 20.25
20.25 = (t + 1)²
±4.5 = t + 1
t = -1 ± 4.5
t = -5.5 or 3.5
The zeros are (-5.5, 0) and (3.5, 0).
Answer:
The answer is D
Step-by-step explanation:
All you have to do is distribute.
x^2(5x-4)+4(5x-4) =
5x^3-4x^2+20x-16
Answer:
Step-by-step explanation:
i'll do a few of them
1) one inch plus a half inch plus an eighth inch plus a sixteenth inch
1 + 1/2 + 1/8 + 1/16 read the tick marks
1 + 8/16 + 2/16 + 1/16 find a common denominators
1 and 11/16 add the sixteenth numerator
2)
3)
4) 6 inch plus an eighth inch plus a sixteenth inch
6 + 1/8 + 1/16 read the tick marks
6 + 2/16 + 1/16 find a common denominators
6 and 3/16 add the sixteenth numerator
5) nine inch plus a quarter inch plus an eighth inch
9 + 1/4 + 1/8 read the tick marks
9 + 2/8 + 1/8 find a common denominators
9 and 3/8 add the sixteenth numerator
