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:
Question 2
x is approximately 15.038 or 15 days
Question 3
$3.41
Question 4
31.1
Step-by-step explanation:
Question 2
5.2 km/1 day =78.2 km/x days
Multiply both sides by x
5.2x=78.2
divide by 5.2
x is approximately 15.038
or 15 days
Question 3
350-(262.94+83.65)
$3.41
Question 4
22(1.3)+2.5
31.1 m
Answer:
Solid dot on -7 with the line going to the right towards 0
Step-by-step explanation:
Some basic formulas involving triangles
\ a^2 = b^2 + c^2 - 2bc \textrm{ cos } \alphaa 2 =b 2+2 + c 2
−2bc cos α
\ b^2 = a^2 + c^2 - 2ac \textrm{ cos } \betab 2=
m_b^2 = \frac{1}{4}( 2a^2 + 2c^2 - b^2 )m b2 = 41(2a 2 + 2c 2-b 2)
b
Bisector formulas
\ \frac{a}{b} = \frac{m}{n} ba =nm
\ l^2 = ab - mnl 2=ab-mm
A = \frac{1}{2}a\cdot b = \frac{1}{2}c\cdot hA=
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
\iits whatever A = prA=pr with r we denote the radius of the triangle inscribed circle
\ A = \frac{abc}{4R}A=
4R
abc
- R is the radius of the prescribed circle
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
