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The body of a man whose weight is 690 N contains about 5.2 × 10-3m3 of blood. The blood mass density is 1060 kg/m3. (a) find the blood’s weight and (b) express it as a percentage of the body weight.



Answer :

To solve this problem, we'll first find the weight of the blood using its mass density and volume, and then we'll express it as a percentage of the body weight.

Given:
Body weight (W_body) = 690 N
Blood volume (V_blood) = 5.2 × 10^(-3) m^3
Blood mass density (ρ_blood) = 1060 kg/m^3

(a) To find the weight of the blood:
We'll use the formula:
Weight = Mass × gravitational acceleration (W = m × g)

First, we'll find the mass of the blood using its density and volume:
m_blood = ρ_blood × V_blood

Then, we'll find the weight of the blood:
W_blood = m_blood × g

(b) To express the blood weight as a percentage of the body weight:
We'll divide the weight of the blood by the body weight and multiply by 100 to get the percentage.

Let's calculate:

(a) Weight of the blood:
m_blood = ρ_blood × V_blood
= 1060 kg/m^3 × 5.2 × 10^(-3) m^3
= 5.552 kg

W_blood = m_blood × g
= 5.552 kg × 9.8 m/s^2
≈ 54.38 N

(b) Percentage of the body weight:
Percentage = (W_blood / W_body) × 100
= (54.38 N / 690 N) × 100
≈ 7.89%

So, (a) the weight of the blood is approximately 54.38 N, and (b) it is expressed as approximately 7.89% of the body weight.

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