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1 answer:
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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 + a
t²
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
Answer:
(a). The maximum loss of water vapor by the person is
(b). The partial pressure of water vapor is
Explanation:
Given that,
Temperature = 37.0°C
Volume of air = 2.00 L
Density of vapor = 44.0 g/m³
We need to calculate the maximum loss of water vapor by the person
Using formula of density
Put the value into the formula
(b). We need to calculate the partial pressure of water vapor
Using formula of pressure
Put the value into the formula
Hence, (a). The maximum loss of water vapor by the person is
(b). The partial pressure of water vapor is