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:
is the answer 14
Step-by-step explanation:
If you are using a calculator, simply enter 28÷100×50 which will give you 14 as the answer.
hope it helps
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Answer:
angles in linear pair are 90° hope it helps u ♥♥♥♥♥♥♥
Answer:
2
Step-by-step explanation:
So we have the equation:
This is in the format point-slope form, where:
Here, m is the slope.
In our original equation, 2 replaces m.
Therefore, our slope is 2.
Answer:
0.0089 cm
Step-by-step explanation:
89/10 000
=0.0089
