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:
5.04
Step-by-step explanation:
1 can equals to 336 grams
so 15 cans will equal to 336 X 15
which is 5040grams
now covert it into kg by dividing it by 1000
which 5.04
<h3>Answer: 7366.96 dollars</h3>
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Use the compound interest formula:
A = P(1+r/n)^(n*t)
where in this case,
A = 12000 = amount after t years
P = unknown = deposited amount we want to solve for
r = 0.05 = the decimal form of 5% interest
n = 1 = refers to the compounding frequency (annual)
t = 10 = number of years
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Plug all these values into the equation, then solve for P
A = P(1+r/n)^(n*t)
12000 = P(1+0.05/1)^(1*10)
12000 = P(1.05)^(10)
12000 = P(1.62889462677744)
12000 = 1.62889462677744P
1.62889462677744P = 12000
P = 12000/1.62889462677744
P = 7366.95904248911
P = 7366.96
Step-by-step explanation:
solution.
if variable d increases then w reduces
w=k.u ×1/d
=ku/d
therefore w=k.u/d
