I don't know what the tables you have look like but her <span>displacement</span> would be -1m
Percentage error:
1.55 – 1.53 ÷ 1.53
0.02 ÷ 1.53
.013 x 100
1.3 % error
I hope this is right.
The net force on the toy is 0 Newton.
The mass of the toy is 2 kg.
The force F₁ is the tension applied by one of the siblings on the toy is 8N.
Let us assume that the force applied by the other sibling is F₂, which is produced by the acceleration is 4m/s².
We know,
Force = Mass x acceleration
Putting values,
F₂ = 2 x 4
F₂ = 8N.
As we know that the forces are in opposite directions because the toy is being pulled by both the siblings in their respective directions which are opposite to each other.
Hence, the net force F is given by,
F = F₁ - F₂
F = 8-8
F 0N.
Hence, there is no net force on the toy.
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Answer:
the tension in the part of the cord attached to the textbook is 7.4989 N
Explanation:
Given the data in the question;
As illustrated in the image below;
first we determine the value of the acceleration,
along vertical direction; we use the second equation of motion;
y = ut + at²
we substitute;
0 m/s for u, 1.29 m for y, 0.850 s for t,
1.29 = 0×0.850 + ×a×(0.850)²
1.29 = 0.36125a
a = 1.29 / 0.36125
a = 3.5709 m/s²
Now when the text book is moving with acceleration , the dynamic equation will be;
T₁ = m₁a
where m₁ is the mass of the text book ( 2.10 kg )
a is the vertical acceleration ( 3.5709 m/s² )
so we substitute
T₁ = 2.10 × 3.5709
T₁ = 7.4989 N
Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N
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