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gayaneshka [121]
3 years ago
15

If 1.4% of the mass of a human body is calcium, how many kilograms of calcium are there in a 195-pound man?

Physics
2 answers:
Mamont248 [21]3 years ago
8 0

Answer:

1.241 kg

Explanation:

The mass of the man in kg is 195 pound * (1/2.2) kg/pound = 195/2.2 = 88.64 kg

If 1.4% of the mass of a human body is calcium, then the mass of calcium inside the 88.64 kg man is

1.4% * 88.64 = 88.64*1.4/100 = 1.241 kg

So there's 1.241 kg of calcium inside a 195-pound man

Degger [83]3 years ago
3 0
<h2>Answer:</h2>

1.24kg

<h2>Explanation:</h2>

Given;

Mass of the man (M) = 195 pound

<em>First, convert the value to kilograms (kg) as follows;</em>

Note that;

1 pound = 0.453592kg

Therefore,

195 pounds = 195 x 0.453592kg =  88.45kg

<em>Now, since the mass of calcium (say m) in a human body is 1.4% of the mass (M) of the body;</em>

=> m = 1.4% of M

=> m = 1.4% x M ------------------(i)

Where;

M = 88.45kg

<em>Substitute the value of M into equation (i) as follows;</em>

=> m = 1.4 % x 88.45

=> m = (1.4 / 100) x 88.45

<em>Solve for m;</em>

m = 0.014 x 88.45

m = 1.24kg

Therefore, the mass of calcium in the 195-pound man is 1.24kg.

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3 years ago
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
svetoff [14.1K]

Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

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Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

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The weight (W=mg) is balanced by restoring force ks at equilibrium position

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W=kx ⇒ k=\frac{W}{x}

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                            = 72

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Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

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3 years ago
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