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MatroZZZ [7]
2 years ago
14

In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

ight in terms of the total body weight. Use the model to find a) the weight of the heart of a human whose weight is 185 lbs. Answer in units of lbs.
Physics
1 answer:
jekas [21]2 years ago
8 0

Answer:

The weight of heart of a human is 0.93 lbs.

Explanation:

Given that,

Approximately weight of heart is 0.5 % of the total body weight.

Weight of human = 185 lbs

Let the the weight of total body is w and weight of heart is w_{h}.

We need to calculate the weight of heart of a human

Using given data

w_{h}=0.5\times w

Where, h = weight of heart

w = weight of human

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

Hence, The weight of heart of a human is 0.93 lbs.

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Substance X is placed in a container with substance Y. Both substances are fluids. Substance X initially sinks to the bottom of
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Answer: Option (A) is the correct answer.

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Convection is a process in which heat transfers from a hotter substance to a colder substance.

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What is the momentum of a 1.5 × 103 kilogram van that is moving at a velocity of 32 meters/second? A. 46.9 kilogram meters/second B. 4.7 × 103 kilogram meters/second C. 4.85 × 102 kilogram meters/second D. 4.85 × 104 kilogram meters/second
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You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.
Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
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1/f = (8 + 12) / 12 * 8 = 20/96
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5 0
2 years ago
A uniform metre rule of weight 0.9 N is suspended horizontally by two vertical loops of thread A and B placed at 20cm and 30cm f
podryga [215]

Answer:

(a) 29 cm

(b) 43.5 cm

Explanation:

(a) when loop A is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark

T at 70 cm mark

-2 N at x

Taking the sum of the torques about B:

∑τ = Iα

(-0.9 N) (50 cm − 70 cm) + (-2 N) (x − 70 cm) = 0

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2 N (x − 70 cm) = 18 Ncm

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The distance from the center is |50 cm − 79 cm| = 29 cm.

(b) when loop B is slack, there are three forces acting on the metre rule.

-0.9 N at 50 cm mark

T at 20 cm mark

-2 N at x

Taking the sum of the torques about A:

∑τ = Iα

(-0.9 N) (50 cm − 20 cm) + (-2 N) (x − 20 cm) = 0

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2 N (x − 20 cm) = -27 Ncm

x − 20 cm = -13.5 cm

x = 6.5 cm

The distance from the center is |50 cm − 6.5 cm| = 43.5 cm

3 0
2 years ago
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