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Juli2301 [7.4K]
3 years ago
9

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. d)the weight of the heart of a whale whose weight is 2.402 × 105 lbs. Answer in units of lbs.
Physics
1 answer:
hammer [34]3 years ago
8 0

Answer:

1201 lbs

Explanation:

Given that in mammals, the weight of the heart is approximately 0.5% of the total body weight.

Let the weight of the heart of a mammal be H

And the weight of the total body be B

The linear model that can gives the heart weight in terms of the total body weight will be:

H = 0.005B

B.) To find the weight of the heart of a whale whose weight is 2.402 × 105 lbs, substitute the whole weight in the formula.

H = 0.005 × 2.402 × 10^5

H = 1201 lbs

Therefore, the weight of the heart of the whale is 1201 lbs

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the answer is

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A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p
saul85 [17]

Answer:

0.67 s

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The displacement, x, of an SHM is given by

x = A\cos(\omega t)

A is the amplitude and \omega is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or \frac{\pi}{4} radian.

From trigonometry, \sin A =\cos B if A and B are complementary.

At t = 0, x = 3.5

3.5 = A\cos(\omega \times0)

A =3.5

So

x = 3.5\cos(\omega t)

At t = 0.12, x = 1.5

1.5 = 3.5\cos(0.12\omega)

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The period, T, is related to \omega by

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5 0
3 years ago
How do you calculate resistance with coulombs, time and J's?
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Hey user


The energy E in joules (J) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C):

E(J) = V(V) ×<span> Q</span>(C)

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joule = volt × coulomb

or

J = V × C

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