Answer:
d
Step-by-step explanation:
the angle is less than 90 degrees
Answer:
8
Step-by-step explanation:
5V-6W
V=4. W=2.
5(4)-6(2)
20-6(2)
20-12=
8
Well draw a box with 10 squares going up and 20 square is going sideways and then shade and 5/8 of it
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:
(a) 120 square units (underestimate)
(b) 248 square units
Step-by-step explanation:
<u>(a) left sum</u>
See the attachment for a diagram of the areas being summed (in orange). This is the sum of the first 4 table values for f(x), each multiplied by 2 (the width of the rectangle). Quite clearly, the curve is above the rectangle for the entire interval, so the rectangle area underestimates the area under the curve.
left sum = 2(1 + 5 + 17 + 37) = 2(60) = 120 . . . . square units
<u>(b) right sum</u>
The right sum is the sum of the last 4 table values for f(x), each multiplied by 2 (the width of the rectangle). This sum is ...
right sum = 2(5 +17 + 37 +65) = 2(124) = 248 . . . . square units
